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Rd Sharma XII Vol 2 2018 Solutions for Class 12 Science Math Chapter 1 Definite Integrals are provided here with simple step-by-step explanations. These solutions for Definite Integrals are extremely popular among Class 12 Science students for Math Definite Integrals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2018 Book of Class 12 Science Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2018 Solutions. All Rd Sharma XII Vol 2 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 22.106:

Question 1:

dydx+2y=e3x

Answer:

We have,dydx+2y=e3x           .....(1)Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=e3x∴ I.F.=e∫P dx          =e∫2 dx         = e2xMultiplying both sides of (1) by e2x, we gete2x dydx+2y=e2xe3x⇒e2xdydx+2e2xy=e5xIntegrating both sides with respect to x, we gety e2x=∫e5xdx+C⇒y e2x=e5x5+C⇒y=15e3x+Ce-2xHence, y=15e3x+Ce-2x is the required solution.

Page No 22.106:

Question 2:

4dydx+8y=5 e-3x

Answer:

We have,
4dydx+8y=5 e-3x

⇒dydx+2y=54e-3x           .....(1)Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=54e-3x∴ I.F.=e∫P dx          =e∫2 dx         = e2xMultiplying both sides of (1) by e2x, we gete2x dydx+2y=54e2xe-3x⇒e2xdydx+2e2xy=54e-xIntegrating both sides with respect to x, we gety e2x=54∫e-x dx+C⇒y e2x=-54e-x+C⇒y=54e-3x+Ce-2xHence, y=54e-3x+Ce-2x is the required solution.

Page No 22.106:

Question 3:

dydx+2y=6 ex

Answer:

We have, dydx+2y=6ex           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=6ex∴ I.F.=e∫P dx          =e∫2 dx         = e2xMultiplying both sides of 1 by e2x, we gete2x dydx+2y=6e2xex⇒e2xdydx+2e2xy=6e3xIntegrating both sides with respect to x, we gety e2x=6∫e3xdx+C⇒y e2x=6e3x3+C⇒y e2x=2e3x+CHence, y e2x=2e3x+C is the required solution.

Page No 22.106:

Question 4:

dydx+y=e-2x

Answer:

We have, dydx+y=e-2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1Q=e-2x ∴ I.F.=e∫P dx          =e∫1 dx         = exMultiplying both sides of 1 by ex, we getex dydx+y=exe-2x⇒exdydx+exy=e-xIntegrating both sides with respect to x, we gety ex=∫e-xdx+C⇒y ex=-e-x+C⇒y=-e-2x+Ce-xHence, y=-e-2x+Ce-x is the required solution.

Page No 22.106:

Question 5:

xdydx=x+y

Answer:

We have,xdydx=x+y⇒dydx=1+1xy ⇒dydx-1xy=1          .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1x Q=1∴ I.F.=e∫P dx          =e-∫1x dx          =e-log x         =elog 1x         =1xMultiplying both sides of 1 by 1x, we get1x dydx-1xy=1x×1⇒1xdydx-1x2y=1xIntegrating both sides with respect to x, we gety1x=∫1x dx+C⇒yx=log x+CHence, yx=log x+C is the required solution.

Page No 22.106:

Question 6:

dydx+2y=4x

Answer:

We have,dydx+2y=4x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2 Q=4x ∴ I.F.=e∫P dx          =e∫2 dx         = e2xMultiplying both sides of 1 by e2x, we gete2x dydx+2y=e2x4x ⇒e2xdydx+2e2xy=e2x4x Integrating both sides with respect to x, we gety e2x=4∫x e2x dx+C⇒y e2x=4∫xI e2xII dx+C⇒y e2x=4x∫e2x dx-4∫ddxx∫e2x dxdx+C⇒y e2x=4xe2x2-4×12∫e2x dx+C⇒y e2x=2x e2x-4×14e2x+C⇒y e2x=2x e2x-e2x+C⇒y e2x=2x-1e2x+C⇒y=2x-1+Ce-2xHence, y=2x-1+Ce-2x is the required solution.

Page No 22.106:

Question 7:

xdydx+y=x ex

Answer:

We have,xdydx+y=x ex⇒dydx+1xy=ex        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1x Q=ex∴ I.F.=e∫P dx          =e∫1x dx          =elog x         =x         Multiplying both sides of 1 by x, we getxdydx+1xy=x ex⇒xdydx+y=xexIntegrating both sides with respect to x, we getxy=∫x exdx+C⇒xy=∫xI exII dx+C⇒xy=x∫ex dx-∫ddxx∫ex dxdx+C⇒xy=x ex-ex+C⇒xy=x-1ex+C⇒y=x-1xex+CxHence, y=x-1xex+Cx is the required solution.

Page No 22.106:

Question 8:

dydx+4xx2+1y+1x2+12=0

Answer:

We have, dydx+4xx2+1y+1x2+12=0    ⇒dydx+4xx2+1y=-1x2+12       .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=4xx2+1 Q=-1x2+12∴ I.F.=e∫P dx          =e2∫2xx2+1 dx          =e2log x2+1         =x2+12Multiplying both sides of 1 by x2+12, we getx2+12dydx+4xx2+1y=x2+12-1x2+12 ⇒x2+12dydx+4xx2+1y=-1Integrating both sides with respect to x, we getx2+12y=-∫dx+C⇒x2+12y=-x+CHence, x2+12y=-x+C  is the required solution.

Page No 22.106:

Question 9:

xdydx+y=x log x

Answer:

We have,
xdydx+y=x log x
Dividing both sides by x, we get
dydx+yx=log xComparing with dydx+Py=Q, we getP=1xQ=log xNow, I.F.=e∫Pdx=e∫1xdx                         =elogx                         =xSo, the solution is given byy×I.F.=∫Q×I.F. dx+C⇒xy=∫x IIlog xI dx+C⇒xy=log x∫xdx-∫ddxlog x∫x dxdx+C⇒xy=x2 log x2-∫x2dx+C⇒xy=x2 log x2-x24+C⇒4xy=2 x2log x-x2+K    where, K=2C

Page No 22.106:

Question 10:

xdydx-y=x-1 ex

Answer:

We have, xdydx-y=x-1ex⇒dydx-1xy=x-1xex         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1x Q=x-1xex∴ I.F.=e∫P dx          =e-∫1x dx          =e-log x         =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1xx-1xex ⇒1xdydx-1x2y=x-1x2exIntegrating both sides with respect to x, we get1xy=∫1x-1x2ex  dx+C⇒1xy=exx+C⇒y=ex+CxHence, y=ex+Cx is the required solution.

Page No 22.106:

Question 11:

dydx+yx=x3

Answer:

We have, dydx+yx=x3         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1xQ=x3∴ I.F.=e∫P dx          =e∫1x dx          =elog x         =x         Multiplying both sides of 1 by x, we getx dydx+1xy=x x3 ⇒xdydx+y=x4Integrating both sides with respect to x, we getxy=∫x4 dx+C⇒xy=x55+C⇒5xy=x5+5C⇒5xy=x5+K            where, K=5CHence, 5xy=x5+K is the required solution.

Page No 22.106:

Question 12:

dydx+y=sin x

Answer:

We have,dydx+y=sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1Q=sin x ∴ I.F.=e∫P dx          =e∫ dx         = exMultiplying both sides of 1 by ex, we getex dydx+y=exsin x⇒exdydx+exy=exsin x Integrating both sides with respect to x, we gety ex=∫exsin x dx+C⇒y ex=ex2sin x-cos x+C⇒y=Ce-x+12sin x-cos xHence, y=Ce-x+12sin x-cos x is the required solution.

Page No 22.106:

Question 13:

dydx+y=cos x

Answer:

We have,dydx+y=cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1Q=cos x  ∴ I.F.=e∫P dx          =e∫ dx         = exMultiplying both sides of (1) by ex, we getex dydx+y=excos x ⇒exdydx+exy=excos xIntegrating both sides with respect to x, we gety ex=∫excos x dx+C⇒y ex=12∫excos x+sin x+-sin x+cos x dx+C⇒yex=ex2cos x+sin x+C⇒y=12cos x+sin x+Ce-xHence, y=12cos x+sin x+Ce-x is the required solution.

Page No 22.106:

Question 14:

dydx+2y=sin x

Answer:

We have,dydx+2y=sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2andQ=sin x  ∴ I.F.=e∫P dx          =e∫2 dx         = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xsin x ⇒e2xdydx+2e2xy=e2xsin xIntegrating both sides with respect to x, we gety e2x=∫e2xsin x  dx+C⇒y e2x=15∫2e2x2sin x-cos x+e2x2 cos x+sin x dx+CPutting e2x2 sin x-cos x=t⇒2e2x2sin x-cos x+e2x2 cos x+sin x dx=dt∴y e2x=15∫dt+C⇒y e2x=t5+Cy e2x=e2x52sin x-cos x+C⇒y=152sin x-cos x+Ce-2xHence, y=152sin x-cos x+Ce-2x is the required solution.

Page No 22.106:

Question 15:

dydx = y tan x − 2 sin x

Answer:

We have, dydx=y tan x-2sin xdydx-y tan x=-2sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-tan xQ=-2sin x  ∴ I.F.=e∫P dx          =e-∫tan x dx         = e-logsec x=cos xMultiplying both sides of 1 by cos x, we getcos x dydx-y tan x=-2sin x×cos x⇒cos xdydx-ysin x=-sin 2x Integrating both sides with respect to x, we gety cos x=-∫sin 2x dx+C⇒ycos x=cos 2x2+C⇒2y cos x=cos 2x+2C⇒2y cos x=cos 2x+K,      where k=2CHence, 2y cos x=cos 2x+K is the required solution.

Page No 22.106:

Question 16:

1+x2dydx+y=tan-1 x

Answer:

We have, 1+x2dydx+y=tan-1x⇒dydx+y1+x2=tan-1x1+x2        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=11+x2 Q=tan-1x1+x2∴ I.F.=e∫P dx          =e∫11+x2 dx          =etan-1xMultiplying both sides of 1 by etan-1x, we getetan-1x dydx+y1+x2=etan-1x tan-1x1+x2⇒etan-1xdydx+etan-1xy1+x2=etan-1xtan-1x1+x2Integrating both sides with respect to x, we getetan-1xy=∫tan-1x×etan-1x1+x2 dx+C⇒etan-1xy=I+C        .....2Here, I=∫tan-1x×etan-1x1+x2 dxPutting tan-1 x=t, we get⇒11+x2dx=dt∴ I=∫t et dt     =t∫etdt-∫ddtt∫etdtdt     =t et-et     =t-1et     =tan-1x-1etan-1xSubstituting the value of I in 2, we getetan-1xy=tan-1x-1etan-1x+C⇒y=tan-1x-1+Ce-tan-1xHence, y=tan-1x-1+Ce-tan-1x is the required solution.

Page No 22.106:

Question 17:

dydx + y tan x = cos x

Answer:

We have,dydx+y tan x=cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=tan xQ=cos x  ∴ I.F.=e∫P dx          =e∫tan x dx         = elogsec x=sec xMultiplying both sides of 1 by sec x, we getsec xdydx+y tan x=cos x ×sec x⇒sec xdydx+y sec x tan x=1Integrating both sides with respect to x, we gety sec x=∫dx+C⇒y sec x=x+CHence, y sec x=x+C  is the required solution.

Page No 22.106:

Question 18:

dydx + y cot x = x2 cot x + 2x

Answer:

We have,dydx+y cot x=x2cot x+2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cot xQ=x2cot x+2x ∴ I.F.=e∫P dx          =e∫cot x dx         = elogsin x=sin xMultiplying both sides of 1 by sin x, we getsin xdydx+ycot x=sin xx2cot x+2x⇒sin xdydx+ycos x=x2cos x+2x sin x Integrating both sides with respect to x, we gety sin x=∫x2Icos xIIdx+∫2x sin x dx+C⇒y sin x=x2∫cos xdx-∫ddxx2∫cos x dxdx+∫2xsin x  dx+C⇒y sin x=x2sin x-∫2xsin x dx+∫2xsin x dx+C⇒y sin x=x2sin x+CHence, y sin x=x2sin x+C is the required solution.

Page No 22.106:

Question 19:

dydx+y tan x=x2 cos2 x

Answer:

We have,
dydx+y tan x=x2 cos2 x
Comparing with dydx+Py=Q, we getP=tan x Q=x2 cos2 xNow,I.F.=e∫tan x dx =elog sec x=sec xTherefore, solution is given byy×I.F.=∫x2 cos2 x×I.F. dx+C⇒y sec x=∫x2 cos x dx+C⇒y sec x=I+CWhere, I=∫x2IIcos x Idx+C⇒ I=x2∫cos x dx-∫ddxx2∫cos x dxdx⇒ I=x2sin x-2∫x sin x dx⇒ I=x2sin x-2∫xI sin xII dx⇒ I=x2sin x-2x∫sin x dx+2∫ddxx∫sin x dxdx⇒ I=x2sin x+2x cos x -2∫cos x dx⇒ I=x2sin x+2x cos x -2sin x⇒ I=x2sin x+2x cos x -2sin x∴ y sec x=x2sin x+2x cos x-2sin x+C⇒y sec x=x2sin x+2x cos x-2sin x+C

Page No 22.106:

Question 20:

1+x2dydx+y=etan-1 x

Answer:

We have, 1+x2dydx+y=etan-1x⇒dydx+y1+x2=etan-1 x1+x2        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=11+x2Q=etan-1 x1+x2∴ I.F.=e∫P dx          =e∫11+x2 dx          =etan-1xMultiplying both sides of 1 by etan-1x, we getetan-1x dydx+y1+x2=etan-1xetan-1 x1+x2⇒etan-1xdydx+y etan-1x1+x2=etan-1xetan-1x1+x2Integrating both sides with respect to x, we gety etan-1x=∫e2tan-1x1+x2 dx+C⇒y etan-1x=I+C        .....2Here,I=∫e2tan-1x1+x2 dxPutting tan-1 x=t, we get11+x2dx=dt∴ I=∫e2t dt     =e2t2     =e2tan-1x2Putting the value of I in 2, we gety etan-1x=e2tan-1x2+C⇒2y etan-1x=e2tan-1x+2C⇒2y etan-1x=e2tan-1x+k, where k=2CHence, 2y etan-1x=e2tan-1x+k is the required solution.

Page No 22.106:

Question 21:

x dy = (2y + 2x4 + x2) dx

Answer:

We have, x dy=2y+2x4+x2dx⇒dydx=2xy+2x3+x⇒dydx-2xy=2x3+x        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-2xQ=2x3+x∴ I.F.=e∫P dx          =e-∫2x dx          =e-2log x         =1x2Multiplying both sides of 1 by 1x2, we get1x2 dydx-2xy=1x2 2x3+x⇒1x2dydx-2x3y=2x+1xIntegrating both sides with respect to x, we get1x2y=∫2x+1xdx+C⇒1x2y=x2+log x+C⇒y=x4+x2log x+Cx2Hence, y=x4+x2log x+Cx2 is the required solution.

Page No 22.106:

Question 22:

1+y2+x-etan-1ydydx=0

Answer:

We have,1+y2+x-etan-1ydydx=0⇒x-etan-1ydydx=-1+y2⇒dydx=-1+y2x-etan-1y⇒dxdy=-x-etan-1y1+y2⇒dxdy+x1+y2=etan-1 y1+y2        .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=11+y2Q=etan-1 y1+y2 ∴ I.F.=e∫P dy          =e∫11+y2 dy          =etan-1yMultiplying both sides of 1 by etan-1y, we getetan-1y dxdy+x1+y2=etan-1y etan-1 y1+y2⇒etan-1ydxdy+x etan-1x1+y2=e2tan-1 y1+y2Integrating both sides with respect to y, we getx etan-1y=∫e2tan-1 y1+y2 dy+C⇒x etan-1y=I+C        .....2Here,I=∫e2tan-1 y1+y2 dyPutting tan-1 y=t, we get11+y2dy=dt∴ I=∫e2t dt     =e2t2     =e2tan-1y2Putting the value of I in 2, we getx etan-1y=e2tan-1y2+C⇒2x etan-1y=e2tan-1y+2C⇒2x etan-1y=e2tan-1y+k     where k=2CHence, 2x etan-1y=e2tan-1y+k is the required solution.

Page No 22.106:

Question 23:

y2dxdy+x-1y=0

Answer:

We have,y2dxdy+x-1y=0⇒y2dxdy+x=1y ⇒dxdy+1y2x=1y3         .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1y2Q=1y3∴ I.F.=e∫P dy          =e∫1y2dy         = e-1yMultiplying both sides of 1 by e-1y, we get e-1ydxdy+x1y2= e-1y1y3⇒ e-1ydxdy+x1y2 e-1y= e-1y 1y3Integrating both sides with respect to y, we getx e-1y=∫ e-1y1y3dy+C⇒x e-1y=I+C         .....2whereI=∫ e-1y1y3dyPutting t=1y, we getdt=-1y2dy∴ I=-∫ t Ie-tIIdt    =-t∫e-tdt+∫ddtt∫e-tdtdt    =te-t+e-t    =t+1e-t    =1y+1e-1yPutting the value of I in 2, we getx e-1y=1y+1e-1y+C ⇒x=y+1y+Ce1yHence, x=y+1y+Ce1y is the required solution.

Page No 22.106:

Question 24:

2x-10y3dydx+y=0

Answer:

We have, 2x-10y3dydx+y=0⇒2x-10y3dydx=-y ⇒dxdy=-1y2x-10y3  ⇒dxdy+2yx=10y2        .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=2yQ=10y2∴ I.F.=e∫P dy          =e∫2ydy         = e2log y=y2Multiplying both sides of 1 by y2, we get  y2dxdy+2yx=  y2×10y2⇒ y2dxdy+2yxy2=10y4Integrating both sides with respect to y, we getx y2=∫ 10y4 dy+C⇒xy2=2y5+C⇒x=2y3+Cy-2Hence, x=2y3+Cy-2 is the required solution.

Page No 22.106:

Question 25:

(x + tan y) dy = sin 2y dx

Answer:

We have,x+tan ydy=sin 2y dx⇒dxdy=x cosec 2y+12sec2y      ⇒dxdy-x cosec 2y=12sec2y         .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-cosec 2yQ=12sec2y∴ I.F.=e∫P dy          =e-∫cosec 2y dy         = e-12logtan y=1tan yMultiplying both sides of 1 by 1tan y, we get 1tan ydxdy-x cosec 2y=12 1tan y×sec2y⇒ 1tan ydxdy-x cosec 2y1tan y=12 1tan y×sec2y Integrating both sides with respect to y, we get1tan yx=∫ 12 1tan y×sec2y dy+C⇒xtan y=I+C         .....2where I=∫12 1tan y×sec2y dyPutting t=tan y, we getdt=sec2 y dy∴ I=12∫1t×dt    =t    =tan yPutting the value of I in 2, we getxtan y=tan y+C ⇒x=tan y+Ctan yHence, x=tan y+Ctan y is the required solution.

Page No 22.106:

Question 26:

dx + xdy = e−y sec2 y dy

Answer:

We have, dx+x dy=e-ysec2y dy⇒ dx=e-ysec2y dy-x dy⇒ dxdy=e-ysec2y-x⇒ dxdy+x=e-ysec2y     .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1Q=e-ysec2y∴ I.F.=e∫P dy          =e∫dy          =eyMultiplying both sides of 1 by ey, we getey dxdy+x=ey e-ysec2y⇒eydxdy+x ey=sec2yIntegrating both sides with respect to y, we getx ey=∫sec2y dy+C⇒x ey=tan y+CHence, x ey=tan y+C is the required solution.

Page No 22.106:

Question 27:

dydx = y tan x − 2 sin x

Answer:

We have,dydx=y tan x-2sin xdydx-y tan x=-2sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-tan xQ=-2sin x∴ I.F.=e∫P dx          =e-∫tan x dx         = e-logsec x=cos xMultiplying both sides of 1 by cos x, we getcos x dydx-ytan x=-2sin x ×cos x⇒cos xdydx+ysin x=-sin 2xIntegrating both sides with respect to x, we gety cos x=-∫sin 2x dx+C⇒y cos x=cos 2x2+C⇒y cos x=1-2sin2x2+C⇒y cos x=-sin2x+12+C⇒y cos x=-sin2x+K     where k=12+C⇒y=sec x-sin2x+KHence, y=sec x-sin2x+K is the required solution.

Page No 22.106:

Question 28:

dydx + y cos x = sin x cos x

Answer:

We have, dydx+y cos x=sin x cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cos xQ=sin x cos x∴ I.F.=e∫P dx          =e∫cosx dx         = esin xMultiplying both sides of 1 by esin x, we get esin xdydx+y cos x= esin xsin x cos x⇒ esin xdydx+esin xy cos x= esin x sin xcos x Integrating both sides with respect to x, we gety esin x=∫ esin x sin xcos x dx+C⇒y esin x=I+C           .....2whereI=∫esin x sin x cos x dxPutting t=sin x, we getdt=cos x dx∴ I=∫etII tI dt     =t∫etdt-∫ddtt∫etdtdt     =t et-et     =ett-1     = esin xsin x-1Putting the value of I in 2, we gety esin x= esin xsin x-1+C⇒y=sin x-1+Ce-sin x Hence, y=sin x-1+Ce-sin x is the required solution.

Page No 22.106:

Question 29:

Solve the following differential equations:

1+x2dydx-2xy=x2+2x2+1        [CBSE 2005]

Answer:

Given, 1+x2dydx-2xy=x2+2x2+1   
⇒dydx-2x1+x2y=x2+2
This is a linear differential equation.
I.F.=e∫-2x1+x2dx=11+x2
y11+x2=∫x2+2x2+1dx⇒y11+x2=∫1+11+x2dx⇒y11+x2=x+tan-1x+C⇒y=x+tan-1x+C1+x2

Page No 22.106:

Question 30:

sin xdydx+y cos x=2 sin2 x cos x

Answer:

We have,sin xdydx+y cos x=2 sin2 x cos x⇒dydx+y cot x=2sin x cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cot xQ=2sin x cos x∴ I.F.=e∫P dx          =e∫cot x dx         = elogsin x=sin xMultiplying both sides of 1 by sin x, we getsin xdydx+y cot x=sin x×2sin xcos x⇒sin xdydx+y cosx=2sin2 xcos xIntegrating both sides with respect to x, we gety sin x=2∫sin2 x cos x dx+C           .....2Putting sin x=t⇒cos x dx=dtTherefore, 2 becomesy sin x=2∫t2 dt+C⇒y sin x=23t3+C⇒y sin x=23sin3x+CHence, y sin x=23sin3x+C is the required solution.

Page No 22.106:

Question 31:

x2-1dydx+2x+2y=2x+1

Answer:

We have,x2-1dydx+2x+2y=2x+1⇒dydx+2x+2x2-1y=2x+1x2-1  ⇒dydx+2x+2x2-1y=2x-1      .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2x+2x2-1Q=2x-1∴ I.F.=e∫P dx          =e∫2x+2x2-1 dx          =e∫2xx2-1+4∫ 1x2-1dx          =elogx2-1+4×12logx-1x+1          =elogx2-1×x-12x+12          =elogx-13x+1          =x-13x+1Multiplying both sides of 1 by x-13x+1, we getx-13x+1 dydx+2x+2x2-1y=x-13x+1×2x-1⇒x-13x+1dydx-2x+2x-12x+12y=2x-12x+1Integrating both sides with respect to x, we getx-13x+1y=∫2x-12x+1 dx+C⇒x-13x+1y=∫2x+12-4xx+1 dx+C⇒x-13x+1y=∫2x+1-4xx+1 dx+C⇒x-13x+1y=∫2x+1-4x+1-1x+1 dx+C⇒x-13x+1y=∫2x+1-4+4x+1 dx+C⇒x-13x+1y=∫2x-6+8x+1 dx+C⇒x-13x+1y=x2-6x+8log x+1+C⇒y=x+1x-13x2-6x+8log x+1+CHence, y=x+1x-13x2-6x+8logx+1 +C is the required solution.

Page No 22.106:

Question 32:

xdydx+2y=x cos x

Answer:

We have, xdydx+2y=x cos x⇒dydx+2xy=cos xComparing with dydx+Py=Q, we getP=2xQ=cos xNow, I.F.=e∫2xdx =e2log x=x2Solution is given by,y×I.F.=∫cos x×I.F. dx+C⇒yx2=∫x2 cos x dx+C⇒x2y=I+C          .....1Where,I=∫x2IIcos x Idx+C⇒ I=x2∫cos x dx-∫ddxx2∫cos x dxdx⇒ I=x2sin x-2∫x sin x dx⇒ I=x2sin x-2∫xI sin xII dx⇒ I=x2sin x-2x∫sin x dx+2∫ddxx∫sin x dxdx⇒ I=x2sin x+2x cos x -2∫cos x dx⇒ I=x2sin x+2x cos x -2sin x⇒ I=x2sin x+2x cos x -2sin xTherefore 1 becomes∴x2y=x2sin x+2x cos x-2sin x+C

Page No 22.106:

Question 33:

dydx-y=xex

Answer:

We have, dydx-y=x ex        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1 Q=ex∴ I.F.=e∫P dx          =e-∫dx          =e-xMultiplying both sides of 1 by e-x, we gete-xdydx-y=x exe-x⇒e-xdydx-e-xy=xIntegrating both sides with respect to x, we gete-xy=∫x dx+C⇒e-xy=x22+C⇒y=x22+Ce xHence, y=x22+Ce x is the required solution.

Page No 22.106:

Question 34:

dydx+2y=xe4x

Answer:

We have, dydx+2y=xe4x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=xe4x∴ I.F.=e∫P dx          =e∫2dx         = e2xMultiplying both sides of 1 by e2x, we get e2xdydx+2y= e2x×xe4x⇒ e2xdydx+ 2e2xy= xe6xIntegrating both sides with respect to x, we get e2xy=∫e6xII xI dx+C⇒e2xy=x∫e6xdx-∫ddxx∫e6xdxdx+C⇒e2xy=xe6x6-e6x36+C⇒y=xe4x6-e4x36+Ce-2xHence, y=xe4x6-e4x36+Ce-2x is the required solution.

Page No 22.106:

Question 35:

Solve the differential equation x+2y2dydx=y, given that when x = 2, y = 1.

Answer:

 We have, x+2y2dydx=y⇒dxdy=1yx+2y2  ⇒dxdy-1yx=2y        .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-1yQ=2y∴ I.F.=e∫P dy          =e-∫1ydy         = e-log y=1yMultiplying both sides of 1 by 1y, we get1ydxdy-1yx= 1y×2y⇒1ydxdy-1y2x=2Integrating both sides with respect to y, we getx1y=∫ 2dy+C⇒x1y=2y+C⇒x=2y2+Cy        .....2Now,y=1 at x=2∴ 2=2+C⇒C=0Putting the value of C in 2, we getx=2y2Hence, x=2y2 is the required solution.



Page No 22.107:

Question 36:

Find one-parameter families of solution curves of the following differential equations:
(or Solve the following differential equations)
(i) dydx+3y=emx, m is a given real number

(ii) dydx-y=cos 2x

(iii) xdydx-y=x+1e-x

(iv) xdydx+y=x4

(v) x log xdydx+y=log x

(vi) dydx-2xy1+x2=x2+2

(vii) dydx+y cos x=esin x cos x

(viii) x+ydydx=1

(ix) dydxcos2 x=tan x-y

(x) e-y sec2 y dy=dx+x dy

(xi) x log xdydx+y=2 log x

(xii) xdydx+2y=x2 log x

Answer:

i We have,dydx+3y=emx           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=3 Q=emx∴ I.F.=e∫P dx          =e∫3 dx         = e3xMultiplying both sides of (1) by e3x, we get e3x dydx+3y=e3xemx ⇒ e3xdydx+3 e3xy=em+3xIntegrating both sides with respect to x, we getye3x=∫em+3x dx+C    when m+3≠0  ⇒ye3x=em+3xm+3+C⇒y=emxm+3+Ce-3x ye3x=∫e0×x dx+C    when m+3=0 ⇒ye3x=∫dx+C⇒ye3x=x+C⇒y=x+Ce-3xHence, y=emxm+3+Ce-3x, where m+3≠0 and y=x+Ce-3x, where m+3=0  are required solutions.


ii We have,dydx-y=cos 2x           .....(1)Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1Q=cos 2x∴ I.F.=e∫P dx          =e-∫ dx         = e-xMultiplying both sides of (1) by e-x, we gete-x dydx-y=e-xcos 2x ⇒e-xdydx-e-xy=e-xcos 2xIntegrating both sides with respect to x, we gety e-x=∫e-xcos 2x  dx+C ⇒y e-x=I+C           .....(2)Where,I=∫e-xcos 2x  dx           .....(3)⇒I=12e-xsin 2x-12∫-e-xsin 2x dx⇒I=12e-xsin 2x+12∫e-xsin 2x dx⇒I=12e-xsin 2x-14e-xcos 2x-12×12∫-e-x×-cos 2x dx⇒I=12e-xsin 2x-14e-xcos 2x-14∫e-xcos 2x dx⇒I=12e-xsin 2x-14e-xcos 2x-14I       From 3⇒54I=12e-xsin 2x-14e-xcos 2x⇒5I=2e-xsin 2x-e-xcos 2x⇒I=e-x52sin 2x-cos 2x           .....(4)From (2) and (4) we get⇒ye-x=e-x52sin 2x-cos 2x+C⇒y=152sin 2x-cos 2x+CexHence, y=152sin 2x-cos 2x+Cex is the required solution.


iii We have, xdydx-y=x+1e-x⇒dydx-1xy=x+1xe-x         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1xQ=x+1xe-x∴ I.F.=e∫P dx          =e-∫1x dx          =e-log x         =1xMultiplying both sides of 1 by 1x, we get1x dydx-1xy=1xx+1xe-x ⇒1xdydx-1x2y=x+1x2e-xIntegrating both sides with respect to x, we get1xy=∫1x+1x2e-xdx+C        .....2Putting 1xe-x=t⇒-1xe-x-1x2e-xdx=dt⇒1x+1x2e-xdx=-dtTherefore 2 becomes1xy=-∫dt+C⇒1xy=-t+C⇒1xy=-1xe-x+C⇒y=-e-x+CxHence, y=-e-x+Cx is the required solution.


iv We have, xdydx+y=x4⇒dydx+1xy=x3         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1x Q=x3∴ I.F.=e∫P dx          =e∫1x dx          =elog x         =xMultiplying both sides of 1 by x, we getx dydx+1xy=x.x3 ⇒xdydx+y=x4 Integrating both sides with respect to x, we getxy=∫x4 dx+C⇒xy=x55+C⇒y=x45+CxHence, y=x45+Cx is the required solution.


v We have,x log xdydx+y=log xDividing both sides by x log x, we getdydx+yx log x= log xx log x⇒dydx+yx log x=1 x⇒dydx+1x log xy=1 xComparing with dydx+Py=Q, we getP=1x log x Q=1 xNow,I.F.=e∫Pdx=e∫1x log xdx                         =eloglog x                         =log xSo, the solution is given byy×I.F.=∫Q×I.F. dx+C⇒y log x=∫1 x×log x dx+C⇒y log x=log x22+C⇒y=12log x+Clog x


vi We have, dydx-2xy1+x2=x2+2     .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-2x1+x2 Q=x2+2∴ I.F.=e∫P dx          =e-∫2x1+x2 dx          =e-log1+x2          =11+x2Multiplying both sides of 1 by 11+x2, we get11+x2 dydx-2xy1+x2=11+x2x2+2⇒11+x2dydx-2xy1+x22=x2+2x2+1Integrating both sides with respect to x, we get11+x2y=∫x2+2x2+1 dx+C⇒11+x2y=∫x2+1+1x2+1 dx+C⇒11+x2y=∫dx+∫1x2+1 dx+C⇒11+x2y=x+tan-1x +C⇒y=1+x2x+tan-1x +CHence, y=1+x2x+tan-1x +C is the required solution.


vii We have,dydx+y cos x=esin x cos x        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cos x Q=esin x cos x∴ I.F.=e∫P dx          =e∫cos x dx          =esin xMultiplying both sides of 1 by esin x, we getesin x dydx+y cos x=esin x×esin x cos x⇒esin xdydx+yesin xcos x=e2sin x cos xIntegrating both sides with respect to x, we getesin xy=∫e2sin x cos x dx+C⇒esin xy=I+C       .....2Where,I=∫e2sin x cos x dxPutting t=sinx, we getdt=cos x dx∴ I=∫e2t dt     =e2t2     =e2sin x2Putting the value of I in 2, we getesin xy=e2sin x2+C⇒y=esin x2+Ce-sin xHence, y=esin x2+Ce-sin x is the required solution.


viii We have,x+ydydx=1⇒dydx=1x+y⇒dxdy=x+y⇒dxdy-x=y       .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-1Q=y∴ I.F.=e∫P dy          =e∫-1 dy          =e-yMultiplying both sides of (1) by e-y, we gete-y dxdy-x=e-yy⇒e-ydxdy-e-yx=e-yyIntegrating both sides with respect to y, we gete-yx=∫y Ie-yII dy+C⇒e-yx=y∫e-y dy-∫ddyy∫e-y dydy+C⇒e-yx=-ye-y -e-y +C⇒e-yx+ye-y +e-y =C⇒x+y+1e-y =C⇒x+y+1=CeyHence, x+y+1=Cey is the required solution.


ix We have,dydxcos2x=tan x-y⇒dydx+1cos2 xy=tan x sec2 x⇒dydx+y sec2 x=tan x sec2 x        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=sec2 xQ=tan x sec2 x∴ I.F.=e∫P dx          =e∫sec2 x dx          =etan xMultiplying both sides of 1 by etan x, we getetan x dydx+y sec2 x=etan xtan x sec2 x⇒etan xdydx+yetan xsec2 x=etan xtan x sec2 xIntegrating both sides with respect to x, we getetan xy=∫etan xtan x sec2 x dx+C⇒etan xy=I+C        .....2Where,I=∫etan xtan x sec2 x dxPutting t= tan x, we getdt=sec2 x dx∴ I=∫tI etII dt    =t∫et dt-∫ddtt∫et dtdt    =t et-et    =t-1et    =tan x-1etan xPutting the value of I in 2, we getetan xy=tan x-1etan x+C⇒y=tan x-1+Ce-tan xHence, y=tan x-1+Ce-tan x is the required solution.


x We have,e-ysec2y dy=dx+x dy⇒ dx=e-ysec2y dy-x dy⇒ dxdy=e-ysec2y-x⇒ dxdy+x=e-ysec2y     .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1Q=e-ysec2y∴ I.F.=e∫P dy          =e∫dy          =eyMultiplying both sides of 1 by ey, we geteydxdy+x=eye-ysec2y⇒eydxdy+eyx=sec2yIntegrating both sides with respect to y, we geteyx=∫sec2y dy+C⇒eyx=tan y+C⇒x=tan y+Ce-yHence, x=tan y+Ce-y is the required solution.


xi We have,x log xdydx+y=2log xDividing both sides by x log x, we getdydx+yx log x=2 log xx log x⇒dydx+yx log x=2 x⇒dydx+1x log xy=2 xComparing with dydx+Py=Q, we getP=1x log xQ=2 xNow, I.F.=e∫Pdx=e∫1x log xdx                         =eloglog x                         =log xSo, the solution is given by y×I.F.=∫Q×I.F. dx+C⇒y log x=2∫1 x×log x dx+CPutting log x=t⇒1xdx=dt∴y log x=2∫t dt+C⇒y log x=2t22+C⇒y log x=t2+C⇒y log x=log x2+C   ∵ log x=t⇒y=log x+Clog x


xii We have,xdydx+2y=x2log xDividing both sides by x, we getdydx+2yx=x log xComparing with dydx+Py=Q, we getP=2xQ=x log xNow, I.F.=e∫Pdx=e∫2xdx                         =e2logx                         =x2So, the solution is given byy×I.F.=∫Q×I.F. dx+C⇒x2y=∫x3IIlog xI dx+C⇒x2y=log x∫x3 dx-∫ddxlog x∫x3 dxdx+C⇒x2y=x4log x4-∫x34dx+C⇒x2y=x4log x4-x416+C⇒y=x2log x4-x216+Cx2⇒y=x2164log x-1+Cx2

Page No 22.107:

Question 37:

Solve each of the following initial value problems:
(i) y'+y=ex, y0=12

(ii) xdydx-y=log x, y1=0

(iii) dydx+2y=e-2x sin x, y0=0

(iv) xdydx-y=x+1e-x, y1=0

(v) 1+y2 dx+x-e-tan-1y dx=0, y0=0

(vi) dydx+y tan x=2x+x2 tan x, y0=1

(vii) xdydx+y=x cos x+sin x, yπ2=1

(viii) dydx+y cot x=4x cosec x, yπ2=0

(ix) dydx+2y tan x=sin x; y=0 when x=π3

(x) dydx-3y cot x=sin 2x; y=2 when x=π2
(xi) dydx+ycotx=2cosx, yπ2=0     
(xii) dy=cosx2-ycosecxdx
(xiii) tanxdydx=2xtanx+x2-y;tanx≠0 given that y = 0 when x=π2.

Answer:

i We have,y'+y=ex⇒dydx+y=ex           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=1 and Q=ex∴ I.F.=e∫P dx          =e∫1 dx         = exMultiplying both sides of 1 by I.F.=ex, we getex dydx+y=exex⇒exdydx+exy=e2xIntegrating both sides with respect to x, we gety ex=∫e2x dx+C⇒y ex=e2x2+C           .....2Now, y0=12∴ 12 e0=e02+C⇒C=0Putting the value of C in 2, we getyex=e2x2⇒ex=ex2Hence, y=ex2  is the required solution.


ii We have,xdydx-y=log x⇒dydx-yx=log xx           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=log xx∴ I.F.=e∫P dx          =e-∫1x dx          =e-log x          =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1x×logxx⇒1xdydx-1x2y=logxx2Integrating both sides with respect to x, we gety1x=∫1x2II×logxI dx+C⇒yx=log x∫1x2dx-∫ddxlog x∫1x2dxdx+C⇒yx=-log xx+∫1x2dx+C⇒yx=-log xx-1x+C⇒y=-log x-1+Cx           .....2Now, y1=0∴ 0=-0-1+C1⇒C=1Putting the value of C in 2, we gety=-log x-1+x⇒y=x-1-log xHence, y=x-1-log x is the required solution.


iii We have, dydx+2y=e-2xsin x           .....1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=2 and Q=e-2xsin x∴ I.F.=e∫P dx          =e∫2 dx         = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xe-2xsin x⇒e2x dydx+2y=sin xIntegrating both sides with respect to x, we getye2x=∫sin x dx+C⇒ye2x=-cos x+C           .....2Now,y0=0∴ 0×e0=-cos 0+C⇒C=1Putting the value of C in 2, we getye2x=-cos x+1⇒ye2x=1-cos xHence, ye2x=1-cos x is the required solution.


iv We have,xdydx-y=x+1e-x⇒dydx-1xy=x+1xe-x         .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=x+1xe-x∴ I.F.=e∫P dx          =e-∫1x dx          =e-log x         =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1xx+1xe-x ⇒1x dydx-1xy=x+1x2e-xIntegrating both sides with respect to x, we get1xy=∫1x+1x2e-x dx+CPutting 1xe-x=t⇒-1xe-x-1x2e-xdx=dt⇒1x+1x2e-x dx=-dt∴1xy=∫-dt+C⇒yx=-t+C⇒yx=-e-xx+C⇒y=-e-x+Cx         .....2Now, y1=0∴ 0=-e-1+C⇒C=e-1Putting the value of C in 2, we gety=-e-x+xe-1⇒y=xe-1-e-xHence, y=xe-1-e-x is the required solution.


v We have,1+y2dx+x-e-tan-1ydy=0⇒x-e-tan-1ydydx=-1+y2⇒1+y2dxdy=-x-e-tan-1y⇒dxdy+x1+y2=e-tan-1 y1+y2        .....1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=11+y2 and Q=e-tan-1 y1+y2∴ I.F.=e∫P dy          =e∫11+y2 dy          =etan-1yMultiplying both sides of 1 by I.F.=etan-1y, we getetan-1y dxdy+x1+y2=etan-1ye-tan-1 y1+y2⇒etan-1y dxdy+x1+y2=11+y2Integrating both sides with respect to y, we getetan-1yx=∫11+y2 dy+C⇒xetan-1y=tan-1y+C        .....2Now, y0=0∴ 0×e0=0+C⇒C=0Putting the value of C in 2, we getxetan-1y=tan-1y+0⇒xetan-1y=tan-1yHence, xetan-1y=tan-1y  is the required solution.


vi We have,dydx+y tan x=2x+x2tan x      .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=tan x and Q=x2cot x+2x∴ I.F.=e∫P dx          =e∫tan x dx         = elogsec x=sec xMultiplying both sides of 1 by I.F.=sec x, we getsec xdydx+ytan x=sec xx2tan x+2x⇒sec xdydx+ytan x=x2tan x sec x+2x sec xIntegrating both sides with respect to x, we getysec x=∫x2tan xsec x dx+2∫xII sec xI dx  +C⇒y sec x=∫x2tan x sec x dx+2sec x∫x dx-2∫ddxsec x∫x dxdx+C⇒y sec x=∫x2tan x sec x dx+x2sec x-∫x2tan x sec x dx+C⇒y sec x=x2sec x+C⇒y=x2+Ccos x      .....2Now,y0=1∴ 1=0+Ccos 0⇒C=1Putting the value of C in 2, we gety=x2+cos xHence, y=x2+cos x is the required solution.


vii We have,xdydx+y=x cos x+sin x⇒dydx+1xy=cos x+sin xx          .....1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=tan x and Q=x2cot x+2x∴ I.F.=e∫P dx          =e∫1xdx         = elog x=xMultiplying both sides of (1) by I.F.=x, we getxdydx+1xy=xcos x+sin xx⇒xdydx+1xy=x cos x+sin xIntegrating both sides with respect to x, we getxy=∫x cos x dx+∫sin x dx+C⇒xy=x sin x-∫1sin xdx-cos x+C⇒xy=x sin x+cos x-cos x+C⇒xy=x sin x+C          .....2Now, yπ2=1∴ 1×π2=π2sinπ2+C⇒C=0Putting the value of C in 2, we getxy=x sin x⇒y=sin xHence, y=sin x is the required solution.


viii We have, dydx+y cot x=4x cosec x           .....1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=cot x and Q=4x cosec x ∴ I.F.=e∫P dx          =e∫cot x dx         = elogsin x=sin xMultiplying both sides of 1 by I.F.=sin x, we getsin xdydx+y cot x=sin x4x cosec x⇒sin xdydx+y cot x=4xIntegrating both sides with respect to x, we gety sin x=4∫x dx+C⇒y sin x=2x2+C           .....2Now,yπ2=0∴ 0×sinπ2=2π22+C⇒C=-π22Putting the value of C in 2, we gety sin x=2x2-π22Hence, y sin x=2x2-π22 is the required solution.


ix We have,dydx+2y tan x=sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2tan x and Q=sin x∴ I.F.=e∫P dx          =e2∫tan x dx         = e2logsec x=sec2xMultiplying both sides of (1) by I.F.=sec2 x, we getsec2 x dydx+2y tan x=sec2 x×sin x⇒sec2 x dydx+2y tan x=tan x sec xIntegrating both sides with respect to x, we gety sec2 x=∫tan x sec x dx+C⇒y sec2 x=sec x+C           .....2Now, yπ3=0∴ 0secπ32=secπ3+C⇒C=-2Putting the value of C in 2, we gety sec2 x=sec x-2⇒y=cos x-2cos2 xHence, y=cos x-2cos2 x  is the required solution.


x We have, dydx-3y cot x=sin 2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-3cot x and Q=sin 2x∴ I.F.=e∫P dx          =e-3∫cot x dx         = e-3logsin x=cosec3xMultiplying both sides of 1 by I.F.=cosec3x, we getcosec3xdydx-3y cot x=sin 2xcosec3x⇒cosec3xdydx-3y cot x=2cot x cosec xIntegrating both sides with respect to x, we gety cosec3x=2∫cot x cosec x dx+C⇒ ycosec3x=-2cosec x+C⇒y=-2sin2x+Csin3x           .....2Now, yπ2=2∴ 2=-2sin2π2+Csin3 π2⇒C=4Putting the value of C in 2, we gety=-2sin2x+4sin3x⇒y=4sin3x-2sin2xHence, y=4sin3x-2sin2x is the required solution.

xi dydx+ycotx=2cosx, yπ2=0     dydx+ycotx=2cosx       ....1    Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=cotx and Q=2cosx∴ I.F.=e∫P dx          =e∫cotx dx         = elogsinx         =sinxMultiplying both sides of 1 by I.F.=sinx, we getsinxdydx+ycotx=2sinxcosx⇒sinxdydx+ycosx=sin2xIntegrating both sides with respect to x, we getysinx=∫sin2x dx+C⇒ysinx=-cos2x2+C           .....2Now, yπ2=0 ∴ 0×sinπ2=-cosπ2+C⇒C=-12Putting the value of C in 2, we getysinx=-cos2x2-12⇒2ysinx=-1+cos2x⇒2ysinx=-2cos2x⇒y=-cotxcosxHence, y=-cotxcosx  is the required solution.


xiidy=cosx2-ycosecxdx⇒dydx=2cosx-ycotx    ⇒dydx+ycotx=2cosx       ....1    Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=cotx and Q=2cosx∴ I.F.=e∫P dx          =e∫cotx dx         = elogsinx         =sinxMultiplying both sides of 1 by I.F.=sinx, we getsinxdydx+ycotx=2sinxcosx⇒sinxdydx+ycosx=sin2xIntegrating both sides with respect to x, we getysinx=∫sin2x dx+C⇒ysinx=-cos2x2+C     Hence, ysinx=-cos2x2+C  is the required solution.

(xiii) 
tanxdydx=2xtanx+x2-y⇒dydx+1tanxy=2xtanx+x2tanx⇒dydx+cotxy=2x+x2cotx
This is a linear differential equation of the form dydx+Py=Q.

Integrating factor, I.F. = e∫Pdx=e∫cotxdx=elogsinx=sinx

The solution of the given differential equation is given by

y×I.F.=∫Q×I.F.dx+C⇒y×sinx=∫2x+x2cotxsinxdx+C⇒ysinx=∫2xsinxdx+∫x2cosxdx+C⇒ysinx=∫2xsinxdx+x2∫cosxdx-∫ddxx2×∫cosxdxdx+C
⇒ysinx=∫2xsinxdx+x2sinx-∫2xsinxdx+C⇒ysinx=x2sinx+C⇒y=x2+cosecx×C        .....1 
It is given that, y = 0 when x=π2.

∴0=π22+cosecπ2×C⇒C=-π24
Putting C=-π24 in (1), we get

y=x2-π24cosecx

Hence, y=x2-π24cosecx is the required solution.

Page No 22.107:

Question 38:

Find the general solution of the differential equation xdydx+2y=x2.

Answer:

We have, xdydx+2y=x2  ⇒dydx+2xy=x        .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2x and Q=x. ∴ I.F.=e∫P dx          =e∫2x dx          =e2log x         =x2         Multiplying both sides of 1 by I.F.=x2, we getx2dydx+2xy=x2x ⇒x2dydx+2xy=x3 Integrating both sides with respect to x, we getx2y=∫x3 dx+C⇒x2y=x44+C⇒y=x24+Cx-2Hence, y=x24+Cx-2  is the required solution.

Page No 22.107:

Question 39:

Find the general solution of the differential equation dydx-y=cos x.

Answer:

We have,dydx-y=cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1 and Q=cos x∴ I.F.=e∫P dx          =e-∫ dx         = e-xMultiplying both sides of 1 by I.F.=e-x, we gete-x dydx-y=e-xcos x ⇒e-xdydx-e-xy=e-xcos xIntegrating both sides with respect to x, we getye-x=∫e-xcos x dx+C⇒ye-x=I+C           .....2Here,I=∫e-xcos x dx           .....3⇒I=e-xsin x-∫-e-xsin x dx⇒I=e-xsin x+∫e-xsin x dx⇒I=e-xsin x-e-xcos x-∫-e-x×-cos x dx⇒I=e-xsin x-e-xcos x-∫e-xcos x dx⇒I=e-xsin x-e-xcos x-I   From  3 ⇒2I=e-xsin x-cos x⇒I=e-x2sin x-cos x           .....4From   2 and 4 we get⇒ye-x=e-x2sin x-cos x+C⇒y=12sin x-cos x+CexHence, y=12sin x-cos x+Cex  is the required solution.

Page No 22.107:

Question 40:

Solve the differential equation y+3x2dxdy=x

Answer:

We have, y+3x2dxdy=x⇒dydx=y+3x2x⇒dydx-1xy=3x        .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=3x∴ I.F.=e∫P dx          =e-∫1x dx          =e-log x         =1xMultiplying both sides of (1) by I.F.=1x, we get1x dydx-1xy=1x3x ⇒1xdydx-1x2y=3Integrating both sides with respect to x, we get1xy=3∫dx+C⇒yx=3x+CHence, yx=3x+C  is the required solution.



Page No 22.108:

Question 41:

Find the particular solution of the differential equation dxdy+x cot y=2y+y2 cot y, y ≠ 0 given that x = 0 when y=π2.

Answer:

We have,dxdy+x cot y=2y+y2cot y          .....1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=cot y and Q=2y+y2cot y∴ I.F.=e∫P dy          =e∫cot y dy         = elogsin y=sin yMultiplying both sides of 1 by I.F.=sin y, we getsin ydxdy+x cot y=sin yy2cot y+2y⇒sin ydxdy+x cos y=y2cos y+2y sin yIntegrating both sides with respect to y, we getx sin y=∫y2Icos yIIdy+∫2y sin y  dy+C⇒x sin y=y2∫cos ydy-∫ddyy2∫cos y dydy+∫2y sin y  dy+C⇒x sin y=y2sin y-∫2y sin y dy+∫2ysin y dy+C⇒x sin y=y2sin y+CNow, y=π2 at x=0∴ 0×sin π2=π42sin π2+C⇒C=-π42Putting the value of C, we getx sin y=y2sin y-π42Hence, x sin y=y2sin y-π42 is the required solution.

Page No 22.108:

Question 42:

Solve the following differential equation:

cot-1y+x dy=1+y2 dx

Answer:

The given differential equation is cot-1y+x dy=1+y2 dx.
This differential equation can be written as
dxdy=cot-1y+x1+y2 ⇒dxdy+-11+y2x=cot-1y1+y2 
This is a linear differential equation with P=-11+y2 and Q=cot-1y1+y2.
I.F. =  e-∫11+y2dy=ecot-1y
Multiply the differential equation by integration factor (I.F.), we get
dxdyecot-1y-x1+y2ecot-1y=cot-1y1+y2ecot-1y ⇒ddyxecot-1y=cot-1y1+y2ecot-1y 
Integrating both sides with respect y, we get
xecot-1y=∫cot-1y1+y2ecot-1y dy+C
Putting t=cot-1y and dt=-11+y2dy, we get
xecot-1y=-∫tet dt+C⇒xecot-1y=-ett-1+C⇒xecot-1y=ecot-1y1-cot-1y+C



Page No 22.134:

Question 1:

The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.

Answer:

Let r be the radius and S be the surface area of the balloon at any time t. Then,
S=4πr2⇒dSdt=8π r drdt         .....1Given: dSdtα t⇒dSdt=kt, where k is any constantPutting dSdt=kt in (1), we get⇒kt=8π r drdtkt dt=8πr drIntegrating both sides, we get ∫kt dt=∫8πr dr⇒kt22=8π ×r22+C        .....(2)At t=0 s, r=1 unit and at t=3 s, r=2 units                   Given∴ 0=8π×12+C⇒C=-4πAnd92k=8π×2+C⇒92k=12 π⇒k=83πSubstituting the values of C and k in (2), we get 8t26π=8π ×r22-4π⇒4t23=4r2-4⇒t23=r2-1⇒r2=1+t23⇒r=1+13t2

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Question 2:

A population grows at the rate of 5% per year. How long does it take for the population to double?

Answer:

Let P0 be the initial population and P be the population at any time t. Then,
dPdt=5P100⇒dPdt=0.05P

⇒dPP=0.05dt Integrating both sides with respect to t, we get∫dPP=∫0.05dt log P=0.05t +CNow,P=P0 at t=0 ∴ log P0=0+C⇒C=log P0Putting the value of C, we getlog P=0.05t+log P0⇒logPP0=0.05tTo find the time when the population will double, we haveP=2P0∴ log2P0P0=0.05t⇒log 2=0.05t⇒t=log 20.05=20 log 2 years

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Question 3:

The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?

Answer:

Let the original population be N and the population at any time t be P.
Given: dPdtαP
⇒dPdt=aP⇒dPP=adt⇒logP=at+C          .....1Now, P=N at t=0Putting P=N and t=0 in 1, we getlogN=CPutting C=logN in 1, we getlogP=at+logN⇒logPN=at         .....2According to the question,log2NN=25a⇒a=125log2=125×0.6931=0.0277Putting a=0.0277 in 2, we getlogPN=0.0277t           ...3For P=500000 and N=100000:log500000100000=0.0277t⇒t=log 50.0277=1.6090.0277=58.08 years=Approximately 58 years

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Question 4:

In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?

Answer:

Let the bacteria count at any time t be N.Given: dNdtα N⇒dNdt=λN⇒1NdN=λdtIntegrating both sides, we get∫1NdN=∫λdt⇒log N=λt+log C             .....(1)Initially when t=0, then N=100000 Given∴log 100000=0+log C⇒log C=log 100000After 2 hours number increased by 10%Therefore, increased number=1000001+10%=110000Given: t=2, N=110000Putting t=2, N=110000 in (1), we getlog 110000=2λ+log 100000⇒12log 1110=λSubstituting the values of log C and λ in (1), we getlog N=t2log 1110+log 100000            .....(2)Now,Let t=T when N=200000Substituting these values in (2), we get log 200000=T2log  1110+log 100000⇒log 2=T2log 1110⇒T=2log 2log1110∴ The count will reach 200000 in 2log 2log1110 hours.

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Question 5:

If the interest is compounded continuously at 6% per annum, how much worth Rs 1000 will be after 10 years? How long will it take to double Rs 1000?

Answer:

Let P0 be the initial amount and P be the amount at any time t. Then,
dPdt=6P100⇒dPdt=0.06P
⇒dPP=0.06dt Integrating both sides with respect to t, we getlog P=0.06t +CNow, P=P0 at t=0 ∴ log P0=0+C⇒C=log P0Putting the value of C, we getlog P=0.06t +log P0⇒logPP0=0.06t⇒e0.06t=PP0To find the amount after 10 years, we get⇒e0.06×10=PP0⇒e0.6=PP0⇒1.822=PP0⇒P=1.822P0⇒P=1.822×1000=Rs 1822To find the time after which the amount will double, we haveP=2P0∴ log2P0P0=0.06t⇒log 2=0.06t⇒t=0.69310.06=11.55 years

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Question 6:

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.

Answer:

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: dPdtαP
⇒dPdt=aP⇒dPP=adt⇒logP=at+C          .....1Now,P=N at t=0 Putting P=N and t=0 in 1, we getlogN=C Putting C=logN in 1, we getlogP=at+logN⇒logPN=at         .....2According to the question,log3NN=5a⇒a=15log3=15×1.0986=0.21972Putting a=0.21972 in 2, we getlogPN=0.21972t           .....3 ⇒e0.21972t=PN             .....4Putting t=10 in 4 to find the bacteria after 10 hours, we get e0.21972×10=PN⇒e2.1972=PN⇒PN=9⇒P=9NTo find the time taken when the number of bacteria becomes 10 times of the number of initial population, we haveP=10N∴ log10NN=15tlog 3⇒t=5 log 10log 3

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Question 7:

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?

Answer:

Let the population at any time t be P.

Given: dPdt α P

⇒dPdt=βP⇒dPP=βdt⇒logP=βt+log C          ...1Now, At t=1990, P=200000 and at t=2000, P=250000∴ log 200000=1990β+log C       ...2      log 250000=2000β+log C        ...3Subtracting 3 from 2, we getlog 200000-log 250000=10β⇒β=110log54Putting β=110log 54 in 2, we getlog 200000=1990×110log54+log C⇒log 200000=199log54+log C   ⇒log C=log 200000-199log54 Putting β=110log 54, log C=log 200000-199 log54 and t=2010 in 1, we getlogP=110×2010log 54+log 200000-199 log54⇒logP=201 log 54+log 200000-199log54⇒logP=log 54201- log54199+log 200000⇒logP=log 5420145199+log 200000⇒logP=log 542+log 200000⇒logP=log2516×200000⇒logP=log  312500⇒P=312500

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Question 8:

If the marginal cost of manufacturing a certain item is given by C' (x) = dCdx= 2 + 0.15 x. Find the total cost function C (x), given that C (0) = 100.

Answer:

We have,dCdx=2+0.15x⇒dC=2+0.15xdxIntegrating both sides with respect to x, we getC=2x+0.152x2+K          .....1At C0=100, we have100=20+0.15202+K⇒K=100Putting the value of T in 1, we getC=2x+0.152x2+100⇒C=0.075x2+2x+100

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Question 9:

A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.

Answer:

Let P0 be the initial amount and P be the amount at any time t.
We have,
dPdt=8P100⇒dPdt=2P25

⇒dPP=225dtIntegrating both sides with respect to t, we getlog P=225t +C          .....1Now,P=P0 at t=0 ∴ log P0=0+C⇒C=log P0Putting the value of C in 1, we getlog P=225t +log P0⇒logPP0=225t⇒e225t=PP0To find the amount after 1 year, we havee225=PP0⇒e0.08=PP0⇒1.0833=PP0⇒P=1.0833P0Percentage increase =P-P0P0×100%                           =1.0833P0-P0P0×100%                           =0.0833×100%                           =8.33%

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Question 10:

In a simple circuit of resistance R, self inductance L and voltage E, the current i at any time t is given by L didt+ R i = E. If E is constant and initially no current passes through the circuit, prove that i=ER1-e-R/Lt.

Answer:

We have,Ldidt+Ri=E⇒didt+RLi=EL            .....1∴ I.F.=e∫RL dt          =eRLtMultiplying both sides of (1) by I.F.=eRLt, we get eRLt didt+RLi=eRLt×EL⇒eRLtdidt+eRLtRLi=eRLt×ELIntegrating both sides with respect to t, we geteRLti=EL∫eRLt dt+C⇒eRLti=EL×LReRLt +C⇒eRLti=EReRLt +C            .....2Now, i=0 at t=0∴ e0×0=ERe0 +C⇒C=-ERPutting the value of C in 2, we geteRLti=EReRLt -ER⇒i=ER -ERe-RLt⇒i=ER 1-e-RLt

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Question 11:

The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.

Answer:

Let the initial amount of radium be N and the amount of radium present at any time t be P.
Given: dPdtαP
⇒dPdt=-aP, where a>0⇒dPP=-adtIntegrating both sides, we get⇒logP=-at+C          .....1Now, P=N at t=0Putting P=N and t=0 in 1, we getlogN=CPutting C=logN in 1, we getlogP=-at+logN⇒logNP=atAccording to the question, logNN2=at⇒log2=at⇒t=1alog2Here, a is the constant of proportionality.

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Question 12:

Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?

Answer:

Let the original amount of the radium be N and the amount of radium at any time t be P.
Given: dPdtαP
⇒dPdt=-aP⇒dPP=-adtIntegrating both sides, we get⇒logP=-at+C          .....1Now,P=N at t=0Putting P=N and t=0 in 1, we getlogN=CPutting C=logN in 1, we getlogP=-at+logN⇒logPN=-at         .....2According to the question,P=12N at t=1590logN2N=-1590a⇒-log 2=-1590a⇒a=11590log 2Putting a=11590log 2 in 2, we getlogPN=-11590log 2t PN=e-log 21590t       .....3Putting t=1 in 4 to find the bacteria after 1 year, we getPN=0.9996⇒P=0.9996NPercentage of amount disapeared in 1 year =N-PN×100%=N-0.9996NN×100%=0.04%



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Question 13:

The slope of the tangent at a point P (x, y) on a curve is -xy. If the curve passes through the point (3, −4), find the equation of the curve.

Answer:

According to the question,dydx=-xy⇒y dy=-x dx Integrating both sides with respect to x, we get∫y dy=-∫x dx⇒y22=-x22+CSince the curve passes through 3,-4, it satisfies the above equation.∴ -422=-322+C⇒8=-92+C⇒C=252Putting the value of C, we gety22=-x22+252⇒x2+y2=25

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Question 14:

Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation y-xdydx=y2+dydx.

Answer:

We have,
y-xdydx=y2+dydx

⇒dydxx+1=y1-y⇒dyy1-y=dxx+1

Integrating both sides, we get∫dyy1-y=∫dxx+1⇒∫1y+11-ydy=∫dxx+1⇒log y-log 1-y=log x+1+C          .....1Since the curve passes throught the point 2, 2, it satisfies the equation of the curve.⇒log 2-log 1-2=log 2+1+C⇒C=log 23Putting the value of C in 1, we getlog y-log 1-y=log x+1+log 23⇒log y1-y=log 2x+13⇒y1-y=2x+13⇒y1-y=±2x+13⇒y1-y=2x+13  or  y1-y=-2x+13Here, given point 2, 2 does not satisfy y1-y=2x+13But it satisfy y1-y=-2x+13∴y1-y=-2x+13⇒yy-1=2x+13⇒3y=2x+1y-1⇒3y=2xy-2x+2y-2⇒2xy-2x-y-2=0

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Question 15:

Find the equation of the curve passing through the point 1,π4 and tangent at any point of which makes an angle tan−1 yx-cos2yx with x-axis.

Answer:

The slope of the curve is given as dydx=tan θ.
Here,
θ=tan-1yx-cos2yx

∴ dydx=tantan-1yx-cos2yx⇒dydx=yx-cos2yx

Let y=vx⇒dydx=v+xdvdx∴ v+xdvdx=v-cos2v⇒xdvdx=-cos2v⇒sec2 v dv=-1xdxIntegrating both sides with respect to x, we get∫sec2 v dv=-∫1xdx⇒tan v=-log x+C⇒tan yx=-log x+CSince the curve passes through 1, π4, it satisfies the above equation.∴ tan π4=-log 1+C⇒C=1Putting the value of C, we gettan yx=-log x+1⇒tan yx=-log x+log e⇒tan yx=logex

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Question 16:

Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.

Answer:

Let the given curve be y = f(x). Suppose P(x,y) be a point on the curve. Equation of the tangent to the curve at P is
Y - y =dydx(X-x) , where (X, Y) is the arbitrary point on the tangent.
Putting Y=0 we get,
0 - y = dydx(X - x)Therefore, X-x=-ydxdy⇒X=x-ydxdyTherefore, cut off by the tangent on the x-axis = x-ydxdyGiven, x-ydxdy=4yTherefore, -ydxdy=4y - x⇒dxdy=x-4yy⇒dydx=yx-4y   ........(1)this is a homogeneous differential equation.
Putting y = vx and dydx=v+xdvdx in (1) we getv + xdvdx = vxx-4vxTherefore, v+xdvdx=v1-4v⇒xdvdx=v1-4v-v = 4v21-4v⇒1-4vv2dv=4dxx
Integrating on both sides we get,
∫1-4vv2dv=4∫dxxTherefore, ∫dvv2-4∫dvv=4∫dxx⇒-1v-4 log v = 4 logx + log c⇒-1v = 4 logx + log c +4 log v⇒4 log(xv) + log c = -1vputting the value of v we get4 log(x×yx) + log c = -xy⇒4 log(y) + log c = -xy⇒log (y4c) = -xy⇒y4c = e-xy

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Question 17:

Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.

Answer:

According to the question,dydx=y+2x⇒dydx-y=2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1 and Q=2x∴ I.F.=e∫P dx          =e-∫ dx         = e-xMultiplying both sides of 1 by I.F.=e-x, we gete-x dydx-y=e-x2x ⇒e-xdydx-e-xy=e-x2x  Integrating both sides with respect to x, we gety e-x=2∫e-xIIxI  dx+C⇒ye-x=2x∫e-xdx-2∫ddxx∫e-xdxdx+C⇒ye-x=-2xe-x-2e-x+C           .....2Since the curve passes through origin, we have0×e0=-2×0×e0-2e0+C⇒C=2Putting the value of C in 2, we getye-x=-2xe-x-2e-x+2⇒y=-2x-2+2ex⇒y+2x+1=2exDISCLAIMER: In the question it should be ex instead of e2x.

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Question 18:

The tangent at any point (x, y) of a curve makes an angle tan−1(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).

Answer:

The slope of the curve is given as dydx=tan θ.
Here,

θ=tan-1 2x+3y∴ dydx=tantan-1 2x+3y⇒dydx=2x+3y

⇒dydx-3y=2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-3 and Q=2x∴ I.F.=e∫P dx          =e∫-3 dx         = e-3xMultiplying both sides of (1), by I.F.=e-3x, we gete-3x dydx-3y=e-3x.2x⇒e-3x dydx-3y=2xe-3xIntegrating both sides with respect to x, we gety e-3x=2∫xe-3x dx+C⇒y e-3x=2∫xIe-3xII dx+C⇒y e-3x=2x∫e-3x dx-2∫ddxx∫e-3x dxdx+C⇒y e-3x=-2xe-3x3+2×13∫e-3x dx+C⇒y e-3x=-23xe-3x-2×19e-3x+C⇒y e-3x=-23xe-3x-29e-3x+CSince the curve passes through 1, 2, it satisfies the above equation.∴ 2e-3=-23e-3-29e-3+C⇒C=2e-3+23e-3+29e-3⇒C=269e-3Putting the value of C, we gety e-3x=-23x-29e-3x+269e-3

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Question 19:

Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).

Answer:



Portion of the x-axis cut off between the origin and tangent at a point =x−y  dxdy=OT

It is given, OT = 2x

∴  x−y  dxdy=2x−x=ydxdy-∫dxx=∫dyy∴  xy=k

Since the curve passes through the point (1, 2)
⇒ at x = 1 ⇒ y = 2
∴ k = 2
∴ xy = 2

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Question 20:

Find the equation to the curve satisfying x (x + 1) dydx-y = x (x + 1) and passing through (1, 0).

Answer:

We have,xx+1dydx-y=xx+1⇒dydx-yxx+1=1Comparing with dydx+Py=Q, we getP=-1xx+1Q=1Now,I.F.=e-∫1xx+1dx             =e-∫1x-1x+1dx                 =e-logxx+1            =x+1x        So, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒x+1xy=∫x+1x dx +C⇒x+1xy=∫dx+∫1xdx +C⇒x+1xy=x+log x+CSince the curve passes throught the point 1, 0, it satisfies the equation of the curve.⇒1+110=1+log 1+C⇒C=-1Putting the value of C in the equation of the curve, we getx+1xy=x+log x-1⇒y=xx+1x+log x-1

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Question 21:

Find the equation of the curve which passes through the point (3, −4) and has the slope 2yx at any point (x, y) on it.

Answer:

According to the question,
dydx=2yx
⇒12ydy=1xdxIntegrating both sides with respect to x, we get∫12ydy=∫1xdx⇒12log y=log x+CSince the curve passes through 3,-4, it satisfies the above equation.∴ 1 2log -4=log 3+C⇒log 2-log 3=C⇒C=log 23Putting the value of C, we getlog y=2log x+2log 23⇒log y=log 49x2⇒y=±49x2⇒9y-4x2=0 or 9y+4x2=0 The given point does not satisfy the equation 9y-4x2=0.  ∴ 9y+4x2=0

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Question 22:

Find the equation of the curve which passes through the origin and has the slope x + 3y − 1 at any point (x, y) on it.

Answer:

According to the question,
dydx=x+3y-1

⇒dydx-3y=x-1
Comparing with dydx+Py=Q, we getP=-3 Q=x-1Now, I.F.=e-∫3dx =e-3xSo, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒ye-3x=∫x-1e-3x dx +C⇒ye-3x =∫xIe-3xIIdx-∫e-3x dx+C⇒ye-3x=x∫e-3x dx-∫ddxx∫e-3x dxdx-∫e-3x dx+C⇒ye-3x=-13xe-3x+13∫e-3x dx-∫e-3x dx+C⇒ye-3x=-13xe-3x-19e-3x +13e-3x+C⇒y=-13x-19 +13+Ce3x⇒y=-13x +29+Ce3xSince the curve passes throught the origin, it satisfies the equation of the curve.⇒0=-0+29+Ce0⇒C=-29Putting the value of C in the equation of the curve, we gety=-13x +291-e3x⇒y+13x=291-e3x⇒33y+x=21-e3x

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Question 23:

At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.

Answer:

According to the question,

dydx=x+xy⇒dydx=x1+y

⇒11+ydy=x dxIntegrating both sides with respect to x, we get∫11+ydy=∫x dx⇒log 1+y=x22+CSince the curve passes through 0, 1, it satisfies the above equation.∴ log 1+1=02+C⇒C=log 2Putting the value of C, we getlog 1+y=x22+log 2⇒log 1+y2=x22⇒1+y2=ex22⇒y+1=2ex22

Page No 22.135:

Question 24:

A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is y2-2xydydx-x2=0, and hence find the curve.

Answer:

Tangent  at P(x, y) is given by Y - y = dydx(X-x)
If p be the perpendicular from the origin, then
p = xdydx-y1+dydx2=x               (given)⇒x2dydx2-2xydydx + y2=x2 +x2 dydx2⇒y2-2xydydx-x2 =0          Hence proved.Now, y2-2xydydx-x2 =0    ⇒dydx = y2-x22xy⇒2xydydx-y2 =-x2 ⇒2ydydx-y2 x=-x Let y2=v⇒dvdx-vx=-x       
Multiplying by the integrating factor e∫-1xdx=1x
v.1x=∫-x.1xdx + c= -x+c⇒y2x2=-x+c⇒x2 + y2=cx

Page No 22.135:

Question 25:

Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.

Answer:


It is given that the distance between the foot of ordinate of point of contact (A) and point of intersection of tangent with x-axis (T) = 2x

Coordinate  of  T=x−ydxdy, 0AT=x−x−y dxdy=2xEquation  of  tangent, Y−y=dydx(X−x)  ⇒y-0=dydxx−x−y dxdy⇒y dxdy=2x⇒∫dxx=2∫dyy⇒lnx=lny2+lncx=cy2As the circle passes through 1, 2.1=c×22⇒c=14⇒4x=y2
 

Page No 22.135:

Question 26:

The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.

Answer:

Let P (x, y) be any point on the curve. The equation of the normal at P (x, y) to the given curve is given as
Y-y=-1dydxX-x
It is given that the curve passes through the point (3, 0). Then,
0-y=-1dydx3-x⇒-y=-1dydx3-x⇒ydydx=3-x⇒y dy=3-xdx⇒y22=3x-x22+C          .....1Since the curve passes through the point 3, 4, it satisfies the equation.⇒422=33-322+C⇒C=8-9+92⇒C=92-1=72Putting the value of C in 1, we gety22=3x-x22+72⇒y2=6x-x2+7⇒x2+y2-6x-7=0

Page No 22.135:

Question 27:

The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.

Answer:

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: dPdtαP
⇒dPdt=aP⇒dPP=adt⇒log P=at+C          .....1Now, P=N at t=0Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=at+log N⇒log PN=at         .....2According to the question,log 2NN=6a⇒a=16log 2Putting a=16log 2  in 2, we getlog PN=t6log 2      .....3Putting t=18 in 3 to find the bacteria after 18 hours, we getlog PN=186 log 2⇒log PN=3log 2⇒log PN=log 8⇒PN=8⇒P=8N

Page No 22.135:

Question 28:

Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will take for one-half of the original amount of  radium to decompose?

Answer:

Let the original amount of radium be N and the amount of radium at any time t be P.
Given: dPdtαP
⇒dPdt=-aP⇒dPP=-a dtIntegrating both sides, we get⇒log P=-at+C          .....1Now,P=N when t=0 Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=-at+log N⇒log PN=-at         .....2According to the question,P=98.9100N=0.989N at t=25∴log 0.989NN=-25a⇒a=-125log 0.989Putting a=-125log 0.989 in 2, we getlogPN=125log 0.989tTo find the time when the radium becomes half of its quantity, we haveN=12P∴log NN2=125log 0.989t⇒log 2=125log 0.989t  ⇒t=25log 2log 0.989=25×0.69310.01106=1566.68≈1567 approx.

Page No 22.135:

Question 29:

Show that all curves for which the slope at any point (x, y) on it is x2+y22xy are rectangular hyperbola.

Answer:

We have,
dydx=x2+y22xyLet y=vxdydx=v+xdvdx∴ v+xdvdx=x2+v2x22vx2⇒xdvdx=1+v22v-v⇒x dvdx=1+v2-2v22v⇒xdvdx=1-v22v⇒2v1-v2dv=1xdx

Integrating both sides, we get∫2v1-v2dy=∫1xdx⇒-log 1-v2=log x-log C⇒log 1-v2C=-log x⇒1-v2=Cx⇒1-yx2=Cx⇒x2-y2x2=Cx⇒x2-y2=Cx
Thus, x2-y2=Cx is the equation of the rectangular hyperbola.



Page No 22.136:

Question 30:

The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.

Answer:

According to the question,
dydx=x+y
⇒dydx-y=x
Comparing with dydx+Py=Q, we getP=-1Q=xNow, I.F.=e-∫dx =e-xSo, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒ye-x=∫xIe-xIIdx+C⇒ye-x=x∫e-x dx-∫ddxx∫e-x dxdx+C⇒ye-x=-xe-x+∫e-x dx+C⇒ye-x=-xe-x-e-x+CSince the curve passes throught the origin, it satisfies the equation of the curve.⇒0e0=-0e0-e0+C⇒C=1Putting the value of C in the equation of the curve, we getye-x=-xe-x-e-x+1⇒ye-x+xe-x+e-x=1⇒y+x+1e-x=1⇒x+y+1=ex

Page No 22.136:

Question 31:

Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.

Answer:

According to the question,
dydx=x+xy
⇒dydx-xy=x
Comparing with dydx+Py=Q, we getP=-xQ=xNow, I.F.=e-∫xdx =e-x22So, the solution is given byy×I.F.=∫Q×I.F. dx +C⇒ye-x22=∫xe-x22dx+C⇒ye-x22=I+CNow,I=∫xe-x22dxPutting -x22=t, we get-xdx=dt∴ I=-∫etdt⇒I=-et⇒I=-e-x22∴ ye-x22=-e-x22+C Since the curve passes throught the point 0, 1, it satisfies the equation of the curve.⇒1e0=-e0+C⇒C=2Putting the value of C in the equation of the curve, we getye-x22=-e-x22+2⇒y=-1+2ex22

Page No 22.136:

Question 32:

The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).

Answer:

According to the question,
dydx=x2
⇒dy=x2dxIntegrating both sides with respect to x, we get∫dy=∫x2dx⇒y=x33+CSince the curve passes through -1, 1, it satisfies the above equation.∴ 1=-13+C⇒C=1+13⇒C=43Putting the value of C, we gety=x33+43⇒3y=x3+4

Page No 22.136:

Question 33:

Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.

Answer:

According to the question,

ydydx=x⇒y dy=x dxIntegrating both sides with respect to x, we get∫y dy=∫x dx⇒y22=x22+CSince the curve passes through 0, a, it satisfies the above equation.∴ a22=02+C⇒C=a22Putting the value of C, we gety22=x22+a22⇒x2-y2=-a2

Page No 22.136:

Question 34:

The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).

Answer:

Let P(x, y) be any point on the curve. Then slope of the tangent at P is dydx.
It is given that the slope of the tangent at P(x,y) is equal to the ordinate  i.e y. 
Therefore dydx = y
⇒1ydy = dx⇒log y = x + log C⇒log y = log ex + log C⇒y = Cex
Since, the curve passes through (1,1). Therefore, x=1 and y=1 . 
Putting these values in equation obtained above we get,
1=Ce1⇒C=1eputting these values in the equation we get,y=ex-1



Page No 22.137:

Question 1:

Define a differential equation.

Answer:

Differential equation:

An equation containing an independent variable, a dependent variable and differential coefficients of the dependent variable with respect to the independent variable is called a differential equation.
for example: dydx=ex+y

Page No 22.137:

Question 2:

Define order of a differential equation.

Answer:

Order of differential equation:

The order of a differential equation is the order of its highest order derivative that apears in the equation.

example: d2ydx2-4dydx=2y
order of the differential equation is 2.

Page No 22.137:

Question 3:

Define degree of a differential equation.

Answer:

Degree of differential equation:

The degree of a differential equation is the power of the highest order derivative occurring in a differential equation when it is written as a polynomial in differential coefficients.

example: d2ydx22-4dydx=2y
the degree of the given differential equation is 2

Page No 22.137:

Question 4:

Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.

Answer:

We have,y= Cx+5     .....1⇒dydx=CSubstituting the value of C in 1, we gety= dydx×x+5⇒xdydx-y+5=0 Hence, xdydx-y+5=0  is the differential equation representing the family of straight lines y=Cx+5, where C is an arbitary constant. 

Page No 22.137:

Question 5:

Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.

Answer:

We have,x2-y2=C2Differentiating with respect to x, we get2x-2ydydx=0⇒2x=2ydydx⇒x dx=y dy⇒x dx-y dy=0Hence, x dx-y dy=0 is the required differential equation.



Page No 22.138:

Question 6:

Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.

Answer:

We have,xy=C2Differentiating with respect to x, we getxdydx+y=0⇒xdydx=-y⇒x dy=-y dx⇒x dy+y dx=0Hence, x dy+y dx=0 is the required differential equation.

Page No 22.138:

Question 7:

Write the degree of the differential equation a2d2ydx2=1+dydx21/4.

Answer:

 We have,a2d2ydx2=1+dydx21/4a2d2ydx24=1+dydx2Degree of the differential equation is the degree of the highest order derivative.Therefore, the degree must be 4.

Page No 22.138:

Question 8:

Write the order of the differential equation 1+dydx2=7 d2ydx23.

Answer:

1+dydx2=7 d2ydx23The order of a differential equation is the order of its highest order derivatives.Here, the required order is 2.

Page No 22.138:

Question 9:

Write the order and degree of the differential equation y=xdydx+a1+dydx2.

Answer:

We have,y=xdydx+a1+dydx2y-xdydx=a1+dydx2Squaring both sides, we gety-xdydx2=a1+dydx22y2+x2dydx2-2xydydx=a21+dydx2y2+dydx2x2-a2-2xydydx=a2From the above equation, we see that the highest order is 1. So, its order is 1 and the power of the highest order derivative is 2.Thus, it is a differential equation of order 1 and degree 2.

Page No 22.138:

Question 10:

Write the degree of the differential equation d2ydx2+xdydx2=2x2 log d2ydx2.

Answer:

We have,d2ydx2+xdydx2=2x2 log d2ydx2⇒d2ydx2+xdydx2-2x2 log d2ydx2=0Here, we observe that LHS of the differential equation cannot be expressed as a polynomial in dydx. So, its degree is not defined. 

Page No 22.138:

Question 11:

Write the order of the differential equation of the family of circles touching X-axis at the origin.

Answer:




The equation of the family of circles touching x-axis at the origin is
x-02+y-a2=a2x2+y2-2ay=0             .....1Here, a is the parameter.Since this equation contains only one arbitary constant, we differentiate it only once. Differentiating with respect to x, we get2x+2ydydx-2adydx=0a=x+ydydxdydx            .....2Putting the value of a from (2) in (1), we get x2+y2=2yx+ydydxdydxx2-y2dydx=2xySo, this is the required differential equation.Here, order of the differential equation is 1.

Page No 22.138:

Question 12:

Write the order of the differential equation of all non-horizontal lines in a plane.

Answer:

The equation of the non-horizontal lines in a plane isy=mx+c, where m is the slope and c is the intercept on y-axis.Differentiating with respect to x, we getdydx=m⇒d2ydx2=0This is the required differential equation. Here, we observe that the order of the required differential equation is 2.

Page No 22.138:

Question 13:

If sin x is an integrating factor of the differential equation dydx+Py=Q, then write the value of P.

Answer:

It is given that sin x is the integrating factor of the differential equation dydx+Py=Q.∴ e∫Pdx=sin x⇒∫P dx=log sin x⇒∫P dx=∫cot x dx          ∵∫cot x dx=log  sin x+C⇒P=cot x  

Page No 22.138:

Question 14:

Write the order of the differential equation of the family of circles of radius r.

Answer:

ans

Page No 22.138:

Question 15:

Write the order of the differential equation whose solution is y = a cos x + b sin x + c e−x.

Answer:

y=a cos x+b sinx +c e-xHere, we see that there are three arbitary constants. Therefore, we differentiate it three times to get rid of all three arbitrary constants.Hence, the order of the differential equation is 3.

Page No 22.138:

Question 16:

Write the order of the differential equation associated with the primitive y = C1 + C2 ex + C3 e−2x + C4, where C1, C2, C3, C4 are arbitrary constants.

Answer:

y=C1+C2ex+C3e-2x+C4the given equation can be reduced to:y=C1+C2ex+C3(e-2x×ec4)Here, ec4 will be a constant.We have 3 constants as C1, C2 and C3.and a differential equation of order n always contains exactly n essential arbitrary constants.Hence, the order of the required differntial equation is 3.

Page No 22.138:

Question 17:

What is the degree of the following differential equation?
5xdydx2-d2ydx2-6y=log x

Answer:

5xdydx2-d2ydx2-6y=log x
Here, we see that  the highest order derivative is d2ydx2 and its power is 1.
Therefore, the given differential equation is of first degree.

Page No 22.138:

Question 18:

Write the degree of the differential equation dydx4+3xd2ydx2=0.

Answer:

dydx4+3xd2ydx2=0
The highest order derivative is d2ydx2 and its power is 1.
Therefore, the given differential equation is of first degree.

Page No 22.138:

Question 19:

Write the degree of the differential equation x d2ydx23+ydydx4+x3=0.

Answer:

d2ydx23+ydydx4+x3=0

The highest order derivative is d2ydx2 and its power is 3.

Therefore, the degree of given differential equation is 3.

Page No 22.138:

Question 20:

Write the differential equation representing family of curves y = mx, where m is arbitrary constant.

Answer:

We have,y=mx     .....1 Differentiating with respect to x⇒dydx=mSubstituting the value of m=dydx in eq 1 we get ,y=xdydxHence, y=xdydx is the required differential equation .

Page No 22.138:

Question 21:

Write the degree of the differential equation x3d2ydx22+xdydx4=0.

Answer:

x3d2ydx22+xdydx4=0

Here, the highest order derivative is d2ydx2 and its power is 2.

Therefore, degree of given differential equation is 2.

Page No 22.138:

Question 22:

Write the degree of the differential equation 1+dydx3=d2ydx22

Answer:

The degree is 2 as the highest derivative is of order 2.

Page No 22.138:

Question 23:

Write the degree of the differential equation
d2ydx2+3dydx2=x2logd2ydx2

Answer:

The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.



Page No 22.139:

Question 24:

Write the degree of the differential equation
d2ydx22+dydx2=xsindydx

Answer:

The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 22.139:

Question 25:

Write the order and degree of the differential equation d2ydx2+dydx14+x15=0

Answer:

The order is 2 as the highest derivative is 2.
The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 22.139:

Question 26:

The degree ofthe differential equation d2ydx2+edydx=0

Answer:

The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 22.139:

Question 27:

How many arbitrary constants are there in the general solution of the differential equation of order 3.

Answer:

The arbitrary constants in the general solution of the differential equation is equal to the order of the differential equation.
Hence, the number of arbitrary constants in the general solution of the differential equation of order 3 are 3.

Page No 22.139:

Question 28:

Write the order of the differential equation representing the family of curves y = ax + a3.

Answer:

The order of the differential equation is equal to the arbitrary constants present in the general solution of the differential equation.
Hence, the order of the differential equation representing the family of curves y = ax + a3 is 1.

Page No 22.139:

Question 29:

Find the sum of the order and degree of the differential equation y=xdydx3+d2ydx2

Answer:

The order is 2 as the highest derivative is 2.
The degree is 1 as the highest derivative is of order 1.
Hence, the sum of the order and degree of the differential equation y=xdydx3+d2ydx2 is 2 + 1 = 3

Page No 22.139:

Question 30:

Find the solution of the differential equation
x1+y2dx+y1+x2dy=0

Answer:

x1+y2dx+y1+x2dy=0⇒y1+x2dy=-x1+y2dx⇒y1+y2dy=-x1+x2dx⇒∫y1+y2dy=-∫x1+x2dx
Let 1+y2=t2 and 1+x2=p2⇒2ydy=2tdt and 2xdx=2pdp⇒ydy=tdt and xdx=pdpSubstituting in above equation, we get⇒∫dt=-∫dp⇒t=-p+C⇒1+x2+1+y2=C

Page No 22.139:

Question 1:

The integrating factor of the differential equation (x log x) dydx+y=2 log x, is given by
(a) log (log x)

(b) ex

(c) log x

(d) x

Answer:

(c) log x

We have,
(x log x) dydx+y=2 log x
Dividing both sides by x log x, we get
dydx+yxlogx=2 log xxlogx⇒dydx+yxlogx=2 x⇒dydx+1xlogxy=2 xComparing with dydx+Py=Q, we getP=1xlogxQ=2 xNow, I.F.=e∫Pdx=e∫1x log xdx                =eloglog x                =log x

Page No 22.139:

Question 2:

The general solution of the differential equation dydx=yx is
(a) log y = kx

(b) y = kx

(c) xy = k

(d) y = k log x

Answer:

(b) y = kx

We have,
dydx=y x⇒1ydy=1 xdxIntegrating both sides, we get∫1ydy=∫1 xdx⇒log y=log x+log k⇒log y-log x=log k⇒logyx=log k⇒yx= k⇒y= kx

Page No 22.139:

Question 3:

Integrating factor of the differential equation cos xdydx+y sin x = 1, is
(a) sin x
(b) sec x
(c) tan x
(d) cos x

Answer:

(b) sec x

We have,
cos xdydx+y sin x=1
Dividing both sides by cos x, we get
dydx+sin xcos xy=1cos x⇒dydx+tan xy=1cos xComparing with dydx+Py=Q, we getP=tan xQ=1 cos xNow, I.F.=e∫tan xdx=elogsec x                      =sec x

Page No 22.139:

Question 4:

The degree of the differential equation d2ydx22-dydx=y3, is
(a) 1/2
(b) 2
(c) 3
(d) 4

Answer:

(b) 2

We have,
d2ydx22-dydx=y3
The highest order derivative is d2yd2x and its power is 2.Hence, the degree is 2.



Page No 22.140:

Question 5:

The degree of the differential equation 5+dydx25/3=x5d2ydx2, is
(a) 4
(b) 2
(c) 5
(d) 10

Answer:

We have,
5+dydx253=x5 d2yd2xTaking Cube power on both sides, we get5+dydx25=x15 d2yd2x3The highest order  derivative is d2yd2x and its power is 3.Hence, the degree is 3.


Disclaimer: The correct option is not given in the question.

Page No 22.140:

Question 6:

The general solution of the differential equation dydx+y cot x = cosec x, is
(a) x + y sin x = C
(b) x + y cos x = C
(c) y + x (sin x + cos x) = C
(d) y sin x = x + C

Answer:

(d) y sin x = x + C


We have,dydx+y cot x=cosec xdydx+ycot x=cosec xComparing with dydx+Py=Q, we getP=cot x Q=cosec xNow, I.F.=e∫cot x dx=elogsin x                       =sin xSo, the solution is given byysinx=∫sin x×cosec x dx+C⇒y sin x=x +C

Page No 22.140:

Question 7:

The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
(a) y" + y' = 0
(b) y" − ω2 y = 0
(c) y" = −ω2 y
(d) y" + y = 0

Answer:

(c) y" = −ω2 y

We have,
y = A cos ωt + B sin ωt                                            .....(1)
Differentiating both sides of (1) with respect to x, we get
dydt=-Aω sin ωt+Bω cos ωt                                            .....(2)
Differentiating both sides of (2) again with respect to x, we get
d2ydt2=-Aω2 cos ωt-Bω2 sin ωt⇒d2ydt2=-ω2A cos ωt+B sin ωt⇒d2ydt2=-ω2y        Using 1∴y''=-ω2y

Page No 22.140:

Question 8:

The equation of the curve whose slope is given by dydx=2yx; x>0, y>0 and which passes through the point (1, 1) is
(a) x2 = y
(b) y2 = x
(c) x2 = 2y
(d) y2 = 2x

Answer:

(a) x2 = y

We have,
dydx=2yx⇒12×1ydy=1 xdxIntegrating both sides, we get12∫1ydy=∫1 xdx⇒12log y=log x+log C⇒log y12-log x=log C⇒logyx=log C⇒yx=C⇒y=Cx          .....1As 1 passes through (1, 1), we get∴1=CPutting the value of C in 1, we gety=x⇒y=x2

Page No 22.140:

Question 9:

The order of the differential equation whose general solution is given by
y = c1 cos (2x + c2) − (c3 + c4) ax + c5 + c6 sin (x − c7) is
(a) 3
(b) 4
(c) 5
(d) 2

Answer:

(c) 5

The given equation can be reduced to : y=c1cos(2x+c2)-(c)ax×ac5+c6sin(x-c7)where c=c3+c4 and ac5 will be a constant

There are 5 constants(c1, c2, c, c6, c7) in the given differential equation.
Hence, the order of the differential equation is 5.

Page No 22.140:

Question 10:

The solution of the differential equation dydx=ax+gby+f represents a circle when
(a) a = b
(b) a = −b
(c) a = −2b
(d) a = 2b

Answer:

(b) a = −b

We have,
dydx=ax+gby+f⇒by+fdy=ax+gdxIntegrating both sides, we get∫by+fdy=∫ax+gdx⇒by22+fy=ax22+gx+C⇒by22+fy-ax22-gx=C⇒by2+2fy-ax2-2gx-2C=0The above equation represents a circle.Therefore, the coffecients of x2 and y2 must be equal. i.e. -a=b⇒a=-b

Page No 22.140:

Question 11:

The solution of the differential equation dydx+2yx=0 with y(1) = 1 is given by
(a) y=1x2

(b) x=1y2

(c) x=1y

(d) y=1x

Answer:

(a) y=1x2

We have,

dydx+2y x=0⇒dydx=-2y x⇒12×1ydy=-1 xdxIntegrating both sides, we get12∫1ydy=-∫1 xdx⇒12log y=-log x+log C⇒log y12+log x=log C⇒logyx=log C⇒yx=C            .....1As1 satisfies y1=1, we get1=CPutting the value of C in 1, we getyx=1⇒y=1x2

Page No 22.140:

Question 12:

The solution of the differential equation dydx-yx+1x=0 is given by
(a) y = xex + C
(b) x = yex
(c) y = x + C
(d) xy = ex + C

Answer:

(a) y = xex + C

We have,
dydx-yx+1x=0⇒dydx=yx+1x⇒dyy=x+1xdxIntegrating both sides, we get∫dyy=∫x+1xdx⇒∫dyy=∫dx+∫1xdx⇒log y=x+log x+C⇒log y-log x=x+C⇒log yx=x+C⇒yx=ex+C⇒y=xex+C

Page No 22.140:

Question 13:

The order of the differential equation satisfying 1-x4+1-y4=ax2-y2 is
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(a) 1

The order of a differential equation depends on the number of arbitrary constants in it.

Since 1-x4+1-y4=ax2-y2 contains only 1 constant, the order of the differential equation is 1.

Page No 22.140:

Question 14:

The solution of the differential equation y1 y3 = y22 is
(a) x = C1 eC2y + C3
(b) y = C1 eC2x + C3
(c) 2x = C1 eC2y + C3
(d) none of these

Answer:

(b) y = C1 eC2x + C3

y1y3=y22y3y2=y2y1⇒d3ydx3d2ydx2=d2ydx2dydx⇒∫ddxd2ydx2d2ydx2=∫ddxdydxdydx⇒lnd2ydx2=lndydx+ln C4⇒d2ydx2=C4dydx⇒∫ddxdydxdydx=∫C4 dxlndydx=C4x+C5⇒dydx=eC4x+C5∫dy=∫ eC4x+C5 dxy=eC4x+C5C4+C6y=eC4x.eC5C4+C6⇒y=C1eC2x+C3where,C1=eC5C4C4=C2C6=C3

Page No 22.140:

Question 15:

The general solution of the differential equation dydx+y g' (x) = g (x) g' (x), where g (x) is a given function of x, is
(a) g (x) + log {1 + y + g (x)} = C
(b) g (x) + log {1 + y − g (x)} = C
(c) g (x) − log {1 + y − g (x)} = C
(d) none of these

Answer:

(b) g (x) + log {1 + y − g (x)} = C


We have,dydx+y g'x=gxg'x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=g'x and Q=gxg'x. ∴ I.F.=e∫P dx          =e∫g'x dx         = egxMultiplying both sides of (1) by I.F., we get egx dydx+yg'x= egx gxg'x⇒ egxdydx+ egxy g'x= egxgxg'xIntegrating both sides with respect to x, we gety egx=∫egxgxg'x dx+K⇒y egx=I+K where I=∫egxgxg'x dxNow,  I=∫egxgxg'x dxPutting gx=t, we getg'x dx=dt∴ I=∫tIetII dt    =t∫et dt-∫ddxt∫et dtdt    =tet-et    =gxegx-egx∴ y egx=gxegx-egx+K⇒y egx+egx-gxegx=K⇒y+1-gx=Ke-gxTaking log on both sides, we getlogy+1-gx=-gx+log K⇒gx+log1+y-gx=C         Where, C=log K

Page No 22.140:

Question 16:

The solution of the differential equation dydx=1+x+y2+xy2, y0=0 is
(a) y2=expx+x22-1

(b) y2=1+C expx+x22

(c) y = tan (C + x + x2)

(d) y=tanx+x22

Answer:

d y=tanx+x22

We have,
dydx=1+x+y2+xy2⇒dydx=x+1+y2x+1⇒dydx=x+11+y2⇒dy1+y2=x+1dxIntegrating both sides, we get∫dy1+y2=∫x+1dx⇒tan-1 y=x22+x+C          .....1Now, y0=0∴ tan-1 0=02+0+C⇒C=0Putting the value of C in 1, we gettan-1 y=x22+x⇒y=tanx22+x



Page No 22.141:

Question 17:

The differential equation of the ellipse x2a2+y2b2=C is
(a) y"y'+y'y-1x=0

(b) y"y'+y'y+1x=0

(c) y"y'-y'y-1x=0

(d) none of these

Answer:

(a) y"y'+y'y-1x=0

We have,
x2a2+y2b2=C                                    .....1
Differentiating with respect to x, we get
2xa2+2yb2y'=0⇒xa2+yb2y'=0                                  .....2Again differentiating with respect to x, we get⇒1a2+1b2y'2+yb2y''=0             .....3Multiplying throughout by x, we getxa2+xb2y'2+xyb2y''=0                  .....4Subtracting 2 from 4, we get1b2xy'2+xyy''-yy'=0 ⇒xy'2+xyy''-yy'=0Dividing both sides by xyy', we gety'y+y''y'-1x=0⇒y''y'+y'y-1x=0 

Page No 22.141:

Question 18:

Solution of the differential equation dydx+yx = sin x is
(a) x (y + cos x) = sin x + C
(b) x (y − cos x) = sin x + C
(c) x (y + cos x) = cos x + C
(d) none of these

Answer:

(a) x (y + cos x) = sin x + C


We have,dydx+yx=sin x⇒dydx+1xy=sin x          .....1Comparing with dydx+Py=Q, we getP=1x Q=sin xNow,I.F.=e∫1xdx =elogx                     =xTherefore, integration of 1 is given byy×I.F.=∫x2×I.F. dx+C ⇒yx=∫xI sin xIIdx+C⇒yx=x∫sin x dx-∫ddxx∫sin x dxdx+C⇒yx=-x cos x+∫cos x dx+C⇒yx+x cos x=sin x+C⇒xy+cos x=sin x+C

Page No 22.141:

Question 19:

The equation of the curve satisfying the differential equation y (x + y3) dx = x (y3 − x) dy and passing through the point (1, 1) is
(a) y3 − 2x + 3x2 y = 0
(b) y3 + 2x + 3x2 y = 0
(c) y3 + 2x −3x2 y = 0
(d) none of these

Answer:

(c) y3 + 2x −3x2 y = 0

We have,

yx+y3dx=xy3-xdyHere, xy+y4dx=xy3-x2dy⇒xydx+y4dx-xy3dy+x2dy=0⇒xydx+xdy+y3ydx-xdy=0⇒xdxy+x2y3ydx-xdyx2=0                   ⇒xdxy-x2y3xdy-ydxx2=0                    ⇒dxyx2y2-yxdyx=0                               ∵Dividing the whole equation by x3y2⇒dxyx2y2=yxdyx
Integrating both sides we get,
⇒∫dxyx2y2=∫yxdyx⇒-1xy=yx22-c∴-1xy-12yx2-c=0∴1xy+12y2x2+c=0∴y3+2x+2cx2y=0

It is given that the curves passes through (1, 1).
Hence,
y3+2x+2cx2y=013+21+2c11=01+2+2c=02c=-3c=-32
∴ The required curve is y3+2x-2×32x2y=0
∴y3+2x-3x2y=0

Page No 22.141:

Question 20:

The solution of the differential equation 2xdydx-y=3 represents
(a) circles
(b) straight lines
(c) ellipses
(d) parabolas

Answer:

(d) parabolas

We have,
2xdydx-y=3⇒2xdydx=3+y⇒13+ydy=12xdxIntegrating both sides, we get∫13+ydy=12∫1xdx⇒log 3+y=12log x+log C⇒log 3+y-log x12=log C⇒log 3+yx=log C⇒3+yx=C⇒3+y=CxSquaring both sides, we get3+y2=Cx             .....1Thus, 1 represents the equation of parabolas.

Page No 22.141:

Question 21:

The solution of the differential equation xdydx=y+x tanyx, is
(a) sinxy=x+C

(b) sinyx=Cx

(c) sinxy=Cy

(d) sinyx=Cy

Answer:

(b) sinyx=Cx

We have,
xdydx=y+x tanyx⇒dydx=yx+tanyx        .....1Let y=vx⇒dydx=v+xdvdxPutting the above value in 1, we getv+xdvdx=v+tan v⇒xdvdx=tan v⇒dvtan v=dxxIntegrating both sides, we getlog sin v= log x+log C⇒log sin v- log x=log C⇒logsin vx=log C⇒sin vx=C⇒sin v=Cx⇒sinyx=Cx          ∵ y=vx

Page No 22.141:

Question 22:

The differential equation satisfied by ax2 + by2 = 1 is
(a) xyy2 + y12 + yy1 = 0
(b) xyy2 + xy12 − yy1 = 0
(c) xyy2 − xy12 + yy1 = 0
(d) none of these

Answer:

(b) xyy2 + xy12 − yy1 = 0

We have,
ax2 + by2 = 1                                           .....(1)
Differentiating both sides of (1) with respect to x, we get
2ax+2bydydx=0                                      .....2
Differentiating both sides of (2) with respect to x, we get
2a+2bdydx2+2byd2ydx2=0⇒2byd2ydx2+dydx2=-2a⇒yd2ydx2+dydx2=-2a2b⇒yd2ydx2+dydx2=--yxdydx                      Using 2⇒xyd2ydx2+dydx2=ydydx⇒xyd2ydx2+xdydx2=ydydx⇒xyd2ydx2+xdydx2-ydydx=0⇒xyy2+xy12-yy1=0

Page No 22.141:

Question 23:

The differential equation which represents the family of curves y = eCx is
(a) y1 = C2 y
(b) xy1 − ln y = 0
(c) x ln y = yy1
(d) y ln y = xy1

Answer:

(d) y ln y = xy1

We have,
y = eCx
Taking ln on both sides, we get
ln y = Cx ln e
⇒ln y=Cx                                    .....1
Differentiating both sides of (1) with respect to x, we get
1yy1=C
Substituting the value of C in (1), we get
ln y=y1yx⇒y ln y=y1x

Page No 22.141:

Question 24:

Which of the following transformations reduce the differential equation dzdx+zxlog z=zx2log z2 into the form dudx+Px u=Qx
(a) u = log x
(b) u = ez
(c) u = (log z)−1
(d) u = (log z)2

Answer:

(c) u = (log z)−1

Given dzdx+zxlog z=zx2log z2          .....1Let u=log z-1dudx=-1log z2×1z×dzdxdzdx=-zlog z2 dudxSubstituting the value of dzdx from equation (1) we get,∴-z log z2 dudx+zxlog z=zx2log z2dudx-1x1log z=-1x2dudx-1xlog z-1=-1x2dudx-1xu=-1x2It can be written as,dudx+pxu=Qxwhere, px=-1x             qx=-1x2


The correct option is C.

Page No 22.141:

Question 25:

The solution of the differential equation dydx=yx+ϕyxϕ'yx is
(a) ϕyx=kx

(b) xϕyx=k

(c) ϕyx=ky

(d) yϕyx=k

Answer:

a ϕyx=kx


We have,dydx=yx+ϕyxϕ'yxLet y=vx⇒dydx=v+xdvdx∴ v+xdvdx=v+ϕvϕ'v⇒xdvdx=ϕvϕ'v⇒ϕvϕ'vdv=1xdxIntegrating both sides, we get∫ϕ'vϕvdv=∫1xdx⇒log ϕv=log x+log k⇒log ϕyx-log x=log k⇒log ϕyxx=log k⇒ ϕyxx= k⇒ϕyx=kx

Page No 22.141:

Question 26:

If m and n are the order and degree of the differential equation y25+4y23y3+y3=x2-1, then
(a) m = 3, n = 3
(b) m = 3, n = 2
(c) m = 3, n = 5
(d) m = 3, n = 1

Answer:

(b) m = 3, n = 2


We have,y25+4y23y3+y3=x2-1⇒y3y25+4y23+y32=y3x2-1The highest order derivative is y3 and its highest exponent in this equation is 2.Therefore, order is 3 and degree is 2.Hence,  m=3, n=2



Page No 22.142:

Question 27:

The solution of the differential equation dydx+1=ex + y, is
(a) (x + y) ex + y = 0
(b) (x + C) ex + y = 0
(c) (x − C) ex + y = 1
(d) (x − C) ex + y + 1 =0

Answer:

(d) (x − C) ex + y + 1 = 0


We have,dydx+1=ex+yLet x+y=v⇒1+dydx=dvdx⇒dydx+1=dvdx∴ dvdx=ev⇒e-vdv=dxIntegrating both sides, we get-e-v=x-C⇒-1=evx-C⇒x-Cex+y+1=0

Page No 22.142:

Question 28:

The solution of x2 + y2 dydx = 4, is
(a) x2 + y2 = 12x + C
(b) x2 + y2 = 3x + C
(c) x3 + y3 = 3x + C
(d) x3 + y3 = 12x + C

Answer:

(d) x3 + y3 = 12x + C

We have,x2+y2dydx=4⇒y2dydx=4-x2⇒y2dy=4-x2dxIntegrating both sides, we get∫y2dy=∫4-x2dx⇒y33=4x-x33+D⇒y3=12x-x3+3D⇒x3+y3=12x+C, where C=3D

Page No 22.142:

Question 29:

The family of curves in which the sub tangent at any point of a curve is double the abscissae, is given by
(a) x = Cy2
(b) y = Cx2
(c) x2 = Cy2
(d) y = Cx

Answer:

(a) x = Cy2

Subtangent=ydydx

It is given that subtangent at any point of a curve is double of the abscissa.

∴ydydx=2xy=2xdydx∫dxx=2∫dyylnx=2lny+alnx=lny2+lnclnx=lncy2x=cy2

Page No 22.142:

Question 30:

The solution of the differential equation x dx + y dy = x2 y dy − y2 x dx, is
(a) x2 − 1 = C (1 + y2)
(b) x2 + 1 = C (1 − y2)
(c) x3 − 1 = C (1 + y3)
(d) x3 + 1 = C (1 − y3)

Answer:

(a) x2 − 1 = C (1 + y2)


We have,
x dx + y dy = x2y dy − y2x dx

⇒x+xy2dx=x2y-ydy⇒xx2-1dx=y1+y2dy⇒2x2x2-1dx=2y21+y2dyIntegrating both sides, we get12∫2y1+y2dy=12∫2xx2-1dx⇒12log1+y2=12logx2-1-12logC⇒log1+y2=logx2-1-logC⇒log1+y2=logx2-1C⇒1+y2=x2-1C⇒C1+y2=x2-1

Page No 22.142:

Question 31:

The solution of the differential equation (x2 + 1) dydx + (y2 + 1) = 0, is
(a) y = 2 + x2

(b) y=1+x1-x

(c) y = x (x − 1)

(d) y=1-x1+x

Answer:

d  y=1-x1+x


We have,x2+1dydx+(y2+1)=0⇒x2+1dydx=-y2+1⇒1y2+1dy=-1 x2+1dxIntegrating both sides, we get∫1y2+1dy=-∫1 x2+1dx⇒tan-1 y=-tan-1 x+tan-1 C⇒tan-1 y+tan-1 x=tan-1 C⇒tan-1x+y1-xy=tan-1 C⇒x+y1-xy=CDisclaimer: The initial value conditions are not given, so the final answer will be obatined only ifC=1. So, ⇒x+y=1-xy⇒y+xy=1-x⇒y1+x=1-x⇒y=1-x1+x

Page No 22.142:

Question 32:

The differential equation xdydx-y=x2, has the general solution
(a) y − x3 = 2cx
(b) 2y − x3 = cx
(c) 2y + x2 = 2cx
(d) y + x2 = 2cx

Answer:

(b) 2y − x3 = cx


We have,

xdydx-y=x2

⇒dydx-1xy=x2Comparing with dydx+Py=Q, we getP=-1x Q=x2Now,I.F.=e-∫1xdx =e-logx                             =elog1x                             =1xy×I.F=∫x2×I.Fdx+C ⇒y1x=∫x2×1xdx+C⇒y1x=∫xdx+C⇒y1x=x22+C⇒2y-x3=Cx                          

Page No 22.142:

Question 33:

The solution of the differential equation dydx-ky=0, y0=1 approaches to zero when x → ∞, if
(a) k = 0
(b) k > 0
(c) k < 0
(d) none of these

Answer:

(c) k < 0


We have,⇒dydx-ky=0⇒dydx=ky⇒1ydy=k dxIntegrating both sides, we get∫1ydy=k∫dx⇒logy=kx+C          .....1Now, y0=1∴ C=0Putting C=0 in 1, we getlogy=kx⇒ekx=yAccording to the question,ek∞=0Since e-∞=0∴ k<0.                          

Page No 22.142:

Question 34:

The solution of the differential equation 1+x2dydx+1+y2=0, is
(a) tan−1 x − tan−1 y = tan−1 C
(b) tan−1 y − tan−1 x = tan−1 C
(c) tan−1 y ± tan−1 x = tan C
(d) tan−1 y + tan−1 x = tan−1 C

Answer:

(d) tan−1y + tan−1x = tan−1C

We have,
1+x2dydx+1+y2=0⇒1+x2dydx=-1+y2⇒11+y2dy=-1 1+x2dxIntegrating both sides we get,∫11+y2dy=-∫1 1+x2dx⇒tan-1y=-tan-1x+tan-1C⇒tan-1y+tan-1x=tan-1C

Page No 22.142:

Question 35:

The solution of the differential equation dydx=x2+xy+y2x2, is
(a) tan-1xy=log y+C

(b) tan-1yx=log x+C

(c) tan-1xy=log x+C

(d) tan-1yx=log y+C

Answer:

b tan-1yx=log x+C

We have,
dydx=x2+xy+y2x2          1
This is homogenous differential equation.
Let y=vx⇒dydx=v+xdvdxNow, putting dydx=v+xdvdx and y=vx in 1, we getv+xdvdx=x2+x2v+x2v2x2⇒v+xdvdx=1+v+v2⇒xdvdx=1+v2⇒11+v2dv=1xdxIntegrating both sides we get,∫11+v2dv=∫1xdx⇒tan-1 v=log x+C⇒tan-1 yx=log x+C

Page No 22.142:

Question 36:

The differential equation dydx+Py=Qyn, n>2 can be reduced to linear form by substituting
(a) z = yn −1
(b) z = yn
(c) z = yn + 1
(d) z = y1 − n

Answer:

(d) z = y1 − n

We have,
dydx+Py=Qyn⇒y-ndydx+Py1-n=Q          .....1Put z=y1-nIntegrating both sides with respect to x, we getdzdx=1-ny-ndydx⇒y-ndydx=11-ndzdxNow, 1 becomes11-ndzdx+Pz=Q⇒dzdx+P1-nz=Q1-nWhich is linear form of differential equation.Therefore, the given differential equation can be reduce to linear form by the substitution,z=y1-n

Page No 22.142:

Question 37:

If p and q are the order and degree of the differential equation ydydx+x3d2ydx2+xy = cos x, then
(a) p < q
(b) p = q
(c) p > q
(d) none of these

Answer:

(c) p > q

We have,
ydydx+x3d2ydx2+xy=cos x
The highest order derivative is d2yd2x and it's degree is 1So, the order is 2 and the degree is 1.∴p=2 and q=1Clearly, p>q



Page No 22.143:

Question 38:

Which of the following is the integrating factor of (x log x) dydx+y = 2 log x?
(a) x
(b) ex
(c) log x
(d) log (log x)

Answer:

(c) log x

We have,
x log xdydx+y=2 log x
Dividing both sides by (x log x) we get,
dydx+yx logx=2 log xx logx⇒dydx+yx logx=2 x⇒dydx+1x logxy=2 xComparing with dydx+Py=Q we get,P=1x logx and Q=2 xNow, I.F=e∫Pdx=e∫1xlogxdx                         =eloglog x                         =log x

Page No 22.143:

Question 39:

What is integrating factor of dydx + y sec x = tan x?
(a) sec x + tan x
(b) log (sec x + tan x)
(c) esec x
(d) sec x

Answer:

(a) sec x + tan x

We have,

dydx+y sec x=tan xComparing with dydx+Py=Q, we getP=sec x Q=tan xNow,I.F.=e∫sec xdx=elogsec x+tan x=sec x+tan x

Page No 22.143:

Question 40:

Integrating factor of the differential equation cos xdydx+y sin x=1, is
(a) cos x
(b) tan x
(c) sec x
(d) sin x

Answer:

(c) sec x

We have,
cos xdydx+y sin x=1
Dividing both sides by cos x, we get
dydx+sin xcos xy=1cos x⇒dydx+tan xy=1 cos xComparing with dydx+Py=Q, we getP=tan xQ=2 cos xNow,I.F.=e∫tan xdx=elogsec x=sec x

Page No 22.143:

Question 41:

The degree of the differential equation d2ydx23+dydx2+sindydx+1=0, is
(a) 3
(b) 2
(c) 1
(d) not defined

Answer:

(d) not defined

We have,
d2ydx23+dydx2+sindydx+1=0
The highest order derivative in this equation is d2yd2x.But the equation cannot be expressed as a polynomial in differential coefficient.Hence, the degree is not defined.

Page No 22.143:

Question 42:

The order of the differential equation 2x2d2ydx2-3dydx+y=0, is
(a) 2
(b) 1
(c) 0
(d) not defined

Answer:

(a) 2

We have,
2x2d2ydx2-3dydx+y=0
Here, the highest order derivative is d2yd2x.Hence, the order is 2.

Page No 22.143:

Question 43:

The number of arbitrary constants in the general solution of differential equation of fourth order is
(a) 0
(b) 2
(c) 3
(d) 4

Answer:

(d) 4

The number of arbitrary constants in the general solution of a differential equation of order n is n.
Thus, the number of arbitrary constants in the general solution of differential equation of fourth order is 4.

Page No 22.143:

Question 44:

The number of arbitrary constants in the particular solution of a differential equation of third order is
(a) 3
(b) 2
(c) 1
(d) 0

Answer:

(d) 0

The number of arbitrary constants in the particular solution of a differential equation is always zero.

Page No 22.143:

Question 45:

Which of the following differential equations has y = C1 ex + C2 e−x as the general solution?
(a) d2ydx2+y=0

(b) d2ydx2-y=0

(c) d2ydx2+1=0

(d) d2ydx2-1=0

Answer:

b  d2ydx2-y=0

We have,
y=C1ex+C2e-x                                    .....1
Differentiating both sides of (1) with respect to x, we get
dydx=C1ex-C2e-x                               .....2
Differentiating both sides of (2) with respect to x, we get
d2ydx2=C1ex+C2e-x⇒d2ydx2=y     Using 1 and 2⇒d2ydx2-y =0

Page No 22.143:

Question 46:

Which of the following differential equations has y = x as one of its particular solution?
(a) d2ydx2-x2dydx+xy=x

(b) d2ydx2+xdydx+xy=x

(c) d2ydx2-x2dydx+xy=0

(d) d2ydx2+xdydx+xy=0

Answer:

c d2ydx2-x2dydx+xy=0

We have,
y = x           .....(1)
Differentiating both sides of (1) with respect to x, we get
dydx=1               .....2Differentiating again with respect to x, we get⇒d2ydx2=0⇒d2ydx2+x2=x2⇒d2ydx2+x×x=x2×1⇒d2ydx2+xy=x2×1          Using 1⇒d2ydx2+xy=x2dydx          Using 2⇒d2ydx2-x2dydx+xy=0                   

Page No 22.143:

Question 47:

The general solution of the differential equation dydx=ex+y, is
(a) ex + e−y = C
(b) ex + ey = C
(c) e−x + ey = C
(d) e−x + e−y = C

Answer:

(a) ex + e−y = C

We have,
dydx=ex+y⇒dydx=ex×ey⇒e-ydy=exdxIntegrating both sides, we get∫e-ydy=∫exdx⇒-e-y=ex+D⇒ex+e-y=-D⇒ex+e-y=C    Where, C=-D

Page No 22.143:

Question 48:

A homogeneous differential equation of the form dxdy=hxy can be solved by making the substitution
(a) y = vx
(b) v = yx
(c) x = vy
(d) x = v

Answer:

(c) x = vy

A homogeneous differential equation of the form dxdy=hxy can be solved by substituting x = vy.

Page No 22.143:

Question 49:

Which of the following is a homogeneous differential equation?
(a) (4x + 6y + 5) dy − (3y + 2x + 4) dx = 0
(b) xy dx − (x3 + y3) dy = 0
(c) (x3 + 2y2) dx + 2xy dy = 0
(d) y2 dx + (x2 − xy − y2) dy = 0

Answer:

(d) y2 dx + (x2 − xy − y2) dy = 0

A differential equation is said to be homogenous if all the terms in the equation have equal degree and it can be written in the form dydx=fx, ygx, y.

In (a), (b) and (c), the degree of all the terms is not equal.
But in the equation y2 dx + (x2 − xy − y2) dy = 0, the degree of all the terms is 2.
Thus, (d) contains a homogeneous differential equation.

Page No 22.143:

Question 50:

The integrating factor of the differential equation xdydx-y=2x2
(a) e−x

(b) e−y

(c) 1x

(d) x

Answer:

c  1x


We have,

xdydx-y=2x2⇒dydx-1xy=2xComparing with dydx+Py=Q, we getP=-1x Q=2xNow,I.F.=e-∫1xdy =e-logx=elog1x=1x                          



Page No 22.144:

Question 51:

The integrating factor of the differential equation 1-y2dxdy+yx=ay-1<y<1 is
(a) 1y2-1

(b) 1y2-1

(c) 11-y2

(d) 11-y2

Answer:

d  11-y2


We have,1-y2dxdy+yx=aydxdy+y1-y2 x=ay1-y2Comparing with dxdy+Px=Q, we getP=y1-y2 Q=ay1-y2Now, I.F.=e∫y1-y2dy=e-12∫-2y1-y2dy=e-12log1-y2=elog11-y2=11-y2

Page No 22.144:

Question 52:

The general solution of the differential equation y dx-x dyy=0, is
(a) xy = C
(b) x = Cy2
(c) y = Cx
(d) y = Cx2

Answer:

(c) y = Cx


We have,y dx-x dyy=0⇒y dx=x dy⇒1ydy=1 xdxIntegrating both sides, we get∫1ydy=∫1 xdx⇒log y=logx+D⇒log y-log x=log C⇒logyx=log C⇒yx=C⇒y=Cx

Page No 22.144:

Question 53:

The general solution of a differential equation of the type dxdy+P1x=Q1 is
(a) ye∫P1dy=∫Q1e∫P1dydy+C

(b) ye∫P1dx=∫Q1e∫P1dxdx+C

(c) xe∫P1dy=∫Q1e∫P1dydy+C

(d) xe∫P1dx=∫Q1e∫P1dxdx+C

Answer:

c  xe∫P1dy=∫Q1e∫P1dydy+C


We have,
dxdy+P1x=Q1
Comparing with the equation dxdy+Px=Q, we get
P = P1
Q = Q1
The general solution of the equation dxdy+Px=Q is given by
xe∫Pdy=∫Qe∫Pdydy+C       ...(1)
Putting the value of P and Q in (1), we get
xe∫P1dy=∫Q1e∫P1dydy+C

Page No 22.144:

Question 54:

The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is
(a) x ey + x2 = C
(b) x ey + y2 = C
(c) y ex + x2 = C
(d) y ey + x2 = C

Answer:

(c) y ex + x2 = C


We have,
ex dy + (yex + 2x) dx = 0
Dividing both sides by exdx, we getdydx+y+2xex=0⇒dydx+y=-2xexComparing with dydx+Py=Q, we getP=1Q=-2xexNow, I.F.=e∫dx =exSolution is given by,                            y×I.F.=∫Q×I.F. dx+C ⇒yex=-∫ex×2xexdx+C⇒yex=-2∫x dx+C⇒yex=-x2+C⇒yex+x2=C              

Page No 22.144:

Question 1:

Determine the order and degree (if defined) of the following differential equations:
(i) dsdt4+3sd2sdt2=0

(ii) y"' + 2y" + y' = 0
(iii) (y"')2 + (y")3 + (y')4 + y5 = 0
(iv) y"' + 2y" + y' = 0
(v) y" + (y')2 + 2y = 0
(vi) y" + 2y' + sin y = 0
(vii) y"' + y2 + ey' = 0

Answer:

i  dsdt4+3sd2sdt2=0
The highest order derivative in the given equation is d2sdt2 and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(ii) y"' + 2y" + y' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1

(iii) (y"')2 + (y")3 + (y')4 + y5 = 0
The highest order derivative in the given equation is y''' and its power is 2.
Therefore, the given differential equation is of third order and second degree.
i.e., Order = 3 and degree = 2

(iv) y"' + 2y" + y' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1

(v) y" + (y')2 + 2y = 0
The highest order derivative in the given equation is y'' and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(vi) y" + 2y' + sin y = 0
The highest order derivative in the given equation is y'' and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(vii) y"' + y2 + ey' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order. This equation cannot be expressed as a polynomial of derivative.
Thus, the degree is not defined.
i.e., Order = 3 and degree is not defined.

Page No 22.144:

Question 2:

Verify that the function y = e−3x is a solution of the differential equation d2ydx2+dydx-6y = 0.

Answer:

We have,d2ydx2+dydx-6y = 0      .....1Now,y=e-3x⇒dydx=-3e-3x⇒d2ydx2=9e-3x
Putting the values of d2ydx2, dydx and y  in 1, we get

LHS=9e-3x-3e-3x-6e-3x      = 0      =RHS

Thus, y = e−3x is the solution of the given differential equation.

Page No 22.144:

Question 3:

In each of the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

(i) y = ex + 1 y'' − y' = 0
(ii) y = x2 + 2x + C y' − 2x − 2 = 0
(iii) y = cos x + C y' + sin x = 0
(iv) y = 1+x2 y' = xy1+x2
(v) y = x sin x xy' = y + x x2-y2
(vi) y=a2-x2 x+ydydx=0

Answer:

(i) We have,
y'' − y' = 0                            .....(1)
Now,
y = ex +1

⇒y'=ex⇒y''=ex

Putting the above values in (1), we get
LHS=ex-ex =0=RHS
Thus, y = ex +1 is the solution of the given differential equation.

(ii) We have,
y' − 2x − 2 = 0                    .....(1)
Now,
y = x2 + 2x + C
⇒y'=2x+2
Putting the above value in (1), we get
LHS=2x+2-2x-2=0=RHS
Thus, y = x2 + 2x + C is the solution of the given differential equation.

(iii) We have,
y' + sin x = 0                    .....(1)
Now,
y = cos x + C
⇒y'=-sin x
Putting the above value in (1), we get
LHS=-sin x+sin x=0=RHS
Thus, y = cos x + C is the solution of the given differential equation.

(iv) We have,
y' = xy1+x2                  .....(1)
Now,
y = 1+x2
⇒y'=x1+x2

Putting the above value in (1), we get
LHS=x1+x2=x1+x2×1+x21+x2=xy1+x2=RHS
Thus, y = 1+x2 is the solution of the given differential equation.

(v) We have,
xy' = y + x x2-y2     .....(1)
Now,
y = x sin x
⇒y'=sin x+x cosx
Putting the above value in (1), we get
LHS=xsin x+xcos x =xsin x+x2cosx=xsin x+xx cos x=xsin x+xx1-sin2 x=xsin x+xx2-x2sin2 x=y+xx2-y2=RHS
Thus, y = x sin x is the solution of the given differential equation.


(v) We have,
xy' = y + x x2-y2         .....(1)
Now,
y = x sin x
⇒y'=sin x+x cosx
Putting the above value in (1), we get
LHS=xsin x+xcos x =xsin x+x2cosx=xsin x+xx cos x=xsin x+xx1-sin2 x=xsin x+xx2-x2sin2 x=y+xx2-y2=RHS
Thus, y = x sin x is the solution of the given differential equation.

(vi) We have,
x+ydydx=0        .....(1)

Now,
y=a2-x2
⇒y'=-xa2-x2

Putting the above value in (1), we get

LHS=x+y-xa2-x2 =x+a2-x2-xa2-x2=x+a2-x2-xa2-x2=x-x=0=RHS                         

Thus, y=a2-x2 is the solution of the given differential equation.



Page No 22.145:

Question 4:

Form the differential equation representing the family of curves y = mx, where m is an arbitrary constant.

Answer:

We have,
y = mx          (1)
Differentiating both sides, we get
dydx=m⇒dydx=yx               From 1⇒xdydx=y⇒xdydx-y=0

Page No 22.145:

Question 5:

Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constant.

Answer:

We have,
y = a sin (x + b)          .....(2)
Differentiating both sides, we get
dydx=a cosx+b⇒d2ydx2=-a sinx+b    ⇒d2ydx2=-a×ya          Using 2⇒d2ydx2=-y ⇒d2ydx2+y=0

Page No 22.145:

Question 6:

Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.

Answer:

The equation of the parabola having vertex at origin and axis along the positive direction of x-axis is given by
y2 =4ax         .....(1)
Since there is only one parameter, so we differentiate it only once.
Differentiating with respect to x, we get
2ydydx=4a
Substituting the value of 4a in (1), we get
y2=2ydydx×x⇒y2=2xydydx⇒y2-2xydydx=0

Page No 22.145:

Question 7:

Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.

Answer:

The equation of the family of circles with radius 3 units, having its centre on y-axis, is given by
x2+y-a2=32                                   .....1
Here, a is any arbitrary constant.
Since this equation has only one arbitrary constant, we get a first order differential equation.
Differentiating (1) with respect to x, we get
2x+2y-adydx=0⇒x+y-adydx=0⇒x=a-ydydx⇒xdydx=a-y⇒a=y+xdydx 
Substituting the value of a in (1), we get

x2+y-y-xdydx2=32⇒x2+x2dydx2=9⇒x2dydx2+x2=9dydx2⇒x2dydx2-9dydx2+x2=0⇒x2-9dydx2+x2=0⇒x2-9y'2+x2=0Hence, x2-9y'2+x2=0  is the required differential equation.

Page No 22.145:

Question 8:

Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Answer:

The equation of the parabola having vertex at origin and axis along the positive direction of y-axis is given by
x2 =4ay         .....(1)
Since there is only one parameter, so we differentiate it only once.
Differentiating with respect to x, we get
2x=4ay'⇒4a=2xy'
Substituting the value of 4a in (1), we get
x2=2xy'×y⇒xy'=2y⇒xy'-2y=0

Page No 22.145:

Question 9:

Form the differential equation of the family of ellipses having foci on y-axis and centre at the origin.

Answer:

The equation of the ellipses having foci on y-axis and centre at the origin is given by
x2a2+y2b2=1            .....(1)
Here,
b > a
Since these are two parameters, so we differentiate the equation twice.
Differentiating with respect to x, we get
2xa2+2yb2y'=0⇒xa2+yb2y'=0                          .....2⇒1a2+1b2y'2+yb2y''=0         .....3Multiplying throughout by x, we getxa2+xb2y'2+xyb2y''=0               .....4Subtracting 2 from 4, we get1b2xy'2+xyy''-yy'=0 ⇒xy'2+xyy''-yy'=0 

Page No 22.145:

Question 10:

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.

Answer:

The equation of the family of hyperbolas having centre at the origin and foci on the X-axis is given by
x2a2-y2b2=1                                        .....1
Here, a and b are parameters.
Since this equation contains two parameters, so we get a second order differential equation.
Differentiating (1) with respect to x, we get
2xa2-2yb2y'=0                                .....2
Differentiating (2) with respect to x, we get
2a2-2b2yy''+y'2=0⇒1a2=1b2yy''+y'2⇒b2a2=yy''+y'2                            .....(3)
From (2), we get
2xa2=2yb2y'⇒b2a2=yxy'                                    .....(4)
From (3) and (4), we get
yxy'=yy''+y'2⇒yy'=xyy''+xy'2Hence, xyy''+xy'2-yy'=0 is the required differential equation.

Page No 22.145:

Question 11:

Verify that xy = a ex + b e−x + x2 is a solution of the differential equation xd2ydx2+2dydx-xy+x2-2=0.

Answer:

We have,xy=aex+be-x+x2Differentiating with respect to x on both sides, we get⇒xdydx+y=aex-be-x+2xAgain differentiating with respect to x on both sides, we get⇒xd2ydx2+dydx+dydx=aex+be-x+2⇒xd2ydx2+2dydx=xy-x2+2      ∵ xy=aex+be-x+x2⇒xd2ydx2+2dydx-xy+x2-2=0
Thus, xy = a ex + b e−x + x2 is the solution of the given differential equation.

Page No 22.145:

Question 12:

Show that y = C x + 2C2 is a solution of the differential equation 2dydx2+xdydx-y=0.

Answer:

We have,
2dydx2+xdydx-y=0               .....1
Now,
y = C x + 2C2
⇒dydx=CPutting dydx=C and y=Cx+2C2 in 1, we get
LHS=2C2+xC-Cx+2C2=2C2+xC-xC-2C2 =0=RHS
 Thus, y = C x + 2C2 is the solution of the given differential equation.

Page No 22.145:

Question 13:

Show that y2 − x2 − xy = a is a solution of the differential equation x-2ydydx+2x+y=0.

Answer:

We have,
x-2ydydx+2x+y=0
Now,y2 − x2 − xy = a
⇒2ydydx-2x-y-xdydx=0⇒2y-xdydx-2x-y=0⇒2y-xdydx=2x+y⇒x-2ydydx=-2x+y⇒x-2ydydx+2x+y=0
 Thus, y2 − x2 − xy = a is the solution of the given differential equation.

Page No 22.145:

Question 14:

Verify that y = A cos x + sin x satisfies the differential equation cos xdydx+sin x y = 1.

Answer:

We have,
 cos xdydx+sin xy=1           .....1
Now,
y = A cos x + sin x
⇒dydx=-A sin x+cos xPutting dydx=-A sin x+cos x and y=A cos x+sin x in 1, we get
LHS=cos x-A sin x+cos x+sin x A cos x+sin x=-A sin x cos x+cos2 x+ A cos x sin x+sin2 x=cos2 x+sin2 x=1=RHS
 Thus, y = A cos x + sin x is the solution of the given differential equation.

Page No 22.145:

Question 15:

Find the differential equation corresponding to y = ae2x + be−3x + cex where a, b, c are arbitrary constants.

Answer:

We have,
y = ae2x + be−3x + cex            .....(1)
Differentiating with respect to x, we get
dydx=2ae2x-3be-3x+cex         .....2⇒d2ydx2=4ae2x+9be-3x+cex⇒d3ydx3=8ae2x-27be-3x+cex⇒d3ydx3=72ae2x-3be-3x+cex-6ae2x+be-3x+cex⇒d3ydx3=7dydx-6y      Using 1 and 2⇒d3ydx3-7dydx+6y=0

Page No 22.145:

Question 16:

Show that the differential equation of all parabolas which have their axes parallel to y-axis is d3ydx3=0.

Answer:

The equation of the family of parabolas axis parallel to y-axis is given by
x-β2=4ay-α                                         .....(1)
Here, α and β are two arbitrary constants.

Differentiating (1) with respect to x, we get
2x-β=4adydx⇒1=2ad2ydx2⇒0=2ad3ydx3⇒d3ydx3=0                                  

Page No 22.145:

Question 17:

From x2 + y2 + 2ax + 2by + c = 0, derive a differential equation not containing a, b and c.

Answer:

We have,
x2 + y2 + 2ax + 2by + c = 0         .....(i)

Differentiating (i) with respect to x, we get

2x+2yy'+2a+2by'=0Again differentiating with respect to x, we get2+2y'2+2yy"+2by"=01+y'2+yy"+by"=0b=-1+y'2+yy"y"We have,1+y'2+yy"+by"=0Again differentiating with respect to x, we get2y'y"+y'y"+yy'''+by'''=0On substituting the value of b we get,3y'y"+yy'''+-1+y'2+yy"y"y'''=03y'y"2+yy"y'''-y'''-y'2y'''-yy'''y"=03y'y"2=y'''1+y'2
 

Page No 22.145:

Question 18:

dydx=sin3 x cos4 x+xx+1

Answer:

We have,dydx=sin3 x cos4 x+xx+1⇒dy=sin3 x cos4 x+xx+1dxIntegrating both sides, we get∫dy=∫sin3 x cos4 x+xx+1dx⇒y=∫sin3 x cos4 x dx +∫xx+1dx ⇒y=I1 +I2           .....1     Here, I1=∫sin3 x cos4 x dxI1=∫xx+1dxNow, I1=∫sin3 x cos4 x dx   =∫1-cos2 x cos4x sin x dxPutting t=cos x, we getdt=-sin x dx∴ I1=-∫t4 1-t2dt       =∫t6-t4dt       =t77-t55+C1       =cos7 x7-cos5 x5+C1 I2=∫xx+1dxPutting t2=x+1, we get2t dt=dx∴I2=2∫t2-1t2dt     =2∫t4-t2 dt      =2t55-2t33+C2      =2x+1525-2x+1323+C2 Putting the value of I1 and I2 in 1, we gety=cos7 x7-cos5 x5+C1+2x+1525-2x+1323+C2 y=cos7 x7-cos5 x5+2x+1525-2x+1323+C     ∵ C=C1+C2Hence, y=cos7 x7-cos5 x5+2x+1525-2x+1323+C   is the solution of the given differential equation.

Page No 22.145:

Question 19:

dydx=1x2+4x+5

Answer:

We have,dydx=1x2+4x+5⇒dydx=1x2+4x+4+1⇒dydx=1x+22+12⇒dy=1x+22+12dxIntegrating both sides, we get∫y=∫1x+22+12dx⇒y=tan-1 x+21+C⇒y=tan-1 x+2+C

Page No 22.145:

Question 20:

dydx=y2+2y+2

Answer:

We have,dydx=y2+2y+2⇒dydx=y2+2y+1+1⇒dydx=y+12+12⇒1y+12+12dy=dxIntegrating both sides, we get∫1y+12+12dy=∫dx⇒tan-1 y+11+C=x⇒x=tan-1y+1+C

Page No 22.145:

Question 21:

dydx+4x=ex

Answer:

We have,dydx+4x=ex⇒dydx=ex-4x⇒dy=ex-4xdxIntegrating both sides, we get∫dy=∫ex-4xdx⇒y=ex-2x2+C⇒y+2x2=ex+C

Page No 22.145:

Question 22:

dydx=x2 ex

Answer:

We have,dydx=x2ex⇒dy=x2ex dxIntegrating both sides, we get∫dy=∫x2IexII dx⇒∫dy=x2∫ex dx-∫ddxx2∫ex dxdx⇒y=x2ex-2∫xex dx⇒y=x2ex-2∫xI exII dx⇒y=x2ex-2x∫ex dx+2∫ddxx∫ex dxdx⇒y=x2ex-2xex+2ex+C⇒y=x2 -2x+2ex+C

Page No 22.145:

Question 23:

dydx-x sin2 x=1x log x

Answer:

We have, dydx-x sin2 x=1x log x⇒dydx=1x log x+x sin2 x⇒dydx=1x log x+x 21-cos 2x⇒dydx=1x log x+x 2-x 2cos 2xIntegrating both sides, we get∫dy=∫1x log x+x 2-x 2cos 2x dx⇒∫dy=∫1x log xdx+12∫x dx-12∫x cos 2xdx⇒∫dy=∫1x log xdx+12∫x dx-12∫xI×cos 2xII dx ⇒y=log log x+x24-x2∫cos 2xdx+12∫ddxx∫cos 2x dxdx⇒y=log log x+x24-x sin 2x4-cos 2x8+C

Page No 22.145:

Question 24:

(tan2 x + 2 tan x + 5) dydx=2 (1 + tan x) sec2 x

Answer:

We have,tan2 x+2 tan x+5dydx=21+tan xsec2 x⇒dy=21+tan xsec2 xtan2 x+2 tan x+5 dxIntegrating both sides, we get∫ dy=∫21+tan xsec2 xtan2 x+2 tan x+5 dx      .....1Putting tan2 x+2 tan x+5=t∴2 tan x sec2x+ 2sec2x dx=dt⇒21+tan xsec2 x dx=dtTherefore 1 becomes,∫ dy=∫1t dt⇒y=log t+C⇒y=log tan2 x+2 tan x+5+C

Page No 22.145:

Question 25:

dydx=sin3 x cos2 x+xex

Answer:

We have,dydx=sin3 x cos2 x+xex⇒dy=sin3 x cos2 x+xexdxIntegrating both sides, we get∫dy=∫sin3 x cos2 x+xexdx⇒y=∫sin3 x cos2 x dx +∫xexdx ⇒y=I1 +I2           .....1     Here, I1=∫sin3x cos2x dxI2=∫xexdxNow, I1=∫sin3 x cos2 x dx=∫1-cos2 x cos2x sin x dxPutting t=cos x, we getdt=-sin x dx∴I1=-∫t2 1-t2dt       =∫-t2+t4dt       =-t33+t55+C1       =cos5 x5-cos3 x3+C1 I2=∫xexdx    =∫xI exII dx    =x∫ex dx-∫ddxx∫ex dxdx    =xex-ex+C2