Rd Sharma XII Vol 2 2020 Solutions for Class 12 Science Maths Chapter 3 Differential Equations are provided here with simple step-by-step explanations. These solutions for Differential Equations are extremely popular among Class 12 Science students for Maths Differential Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2020 Book of Class 12 Science Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2020 Solutions. All Rd Sharma XII Vol 2 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 21.106:

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.=eP dx          =e2 dx         = e2xMultiplying both sides of (1) by e2x, we gete2x dydx+2y=e2xe3xe2xdydx+2e2xy=e5xIntegrating both sides with respect to x, we gety e2x=e5xdx+Cy e2x=e5x5+Cy=15e3x+Ce-2xHence, y=15e3x+Ce-2x is the required solution.

Page No 21.106:

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.=eP dx          =e2 dx         = e2xMultiplying both sides of (1) by e2x, we gete2x dydx+2y=54e2xe-3xe2xdydx+2e2xy=54e-xIntegrating both sides with respect to x, we gety e2x=54e-x dx+Cy e2x=-54e-x+Cy=54e-3x+Ce-2xHence, y=54e-3x+Ce-2x is the required solution.

Page No 21.106:

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.=eP dx          =e2 dx         = e2xMultiplying both sides of 1 by e2x, we gete2x dydx+2y=6e2xexe2xdydx+2e2xy=6e3xIntegrating both sides with respect to x, we gety e2x=6e3xdx+Cy e2x=6e3x3+Cy e2x=2e3x+CHence, y e2x=2e3x+C is the required solution.

Page No 21.106:

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.=eP dx          =e1 dx         = exMultiplying both sides of 1 by ex, we getex dydx+y=exe-2xexdydx+exy=e-xIntegrating both sides with respect to x, we gety ex=e-xdx+Cy ex=-e-x+Cy=-e-2x+Ce-xHence, y=-e-2x+Ce-x is the required solution.

Page No 21.106:

Question 5:

xdydx=x+y

Answer:

We have,xdydx=x+ydydx=1+1xy dydx-1xy=1          .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1x Q=1 I.F.=eP dx          =e-1x dx          =e-log x         =elog 1x         =1xMultiplying both sides of 1 by 1x, we get1x dydx-1xy=1x×11xdydx-1x2y=1xIntegrating both sides with respect to x, we gety1x=1x dx+Cyx=log x+CHence, yx=log x+C is the required solution.

Page No 21.106:

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.=eP dx          =e2 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=4x e2x dx+Cy e2x=4xI e2xII dx+Cy e2x=4xe2x dx-4ddxxe2x dxdx+Cy e2x=4xe2x2-4×12e2x dx+Cy e2x=2x e2x-4×14e2x+Cy e2x=2x e2x-e2x+Cy e2x=2x-1e2x+Cy=2x-1+Ce-2xHence, y=2x-1+Ce-2x is the required solution.

Page No 21.106:

Question 7:

xdydx+y=x ex

Answer:

We have,xdydx+y=x exdydx+1xy=ex        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1x Q=ex I.F.=eP dx          =e1x dx          =elog x         =x         Multiplying both sides of 1 by x, we getxdydx+1xy=x exxdydx+y=xexIntegrating both sides with respect to x, we getxy=x exdx+Cxy=xI exII dx+Cxy=xex dx-ddxxex dxdx+Cxy=x ex-ex+Cxy=x-1ex+Cy=x-1xex+CxHence, y=x-1xex+Cx is the required solution.

Page No 21.106:

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.=eP dx          =e22xx2+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+Cx2+12y=-x+CHence, x2+12y=-x+C  is the required solution.

Page No 21.106:

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.=ePdx=e1xdx                         =elogx                         =xSo, the solution is given byy×I.F.=Q×I.F. dx+Cxy=x IIlog xI dx+Cxy=log xxdx-ddxlog xx dxdx+Cxy=x2 log x2-x2dx+Cxy=x2 log x2-x24+C4xy=2 x2log x-x2+K    where, K=2C

Page No 21.106:

Question 10:

xdydx-y=x-1 ex

Answer:

We have, xdydx-y=x-1exdydx-1xy=x-1xex         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1x Q=x-1xex I.F.=eP 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+C1xy=exx+Cy=ex+CxHence, y=ex+Cx is the required solution.

Page No 21.106:

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.=eP dx          =e1x 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+Cxy=x55+C5xy=x5+5C5xy=x5+K            where, K=5CHence, 5xy=x5+K is the required solution.

Page No 21.106:

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.=eP dx          =e dx         = exMultiplying both sides of 1 by ex, we getex dydx+y=exsin xexdydx+exy=exsin x Integrating both sides with respect to x, we gety ex=exsin x dx+Cy ex=ex2sin x-cos x+Cy=Ce-x+12sin x-cos xHence, y=Ce-x+12sin x-cos x is the required solution.

Page No 21.106:

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.=eP 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+Cy ex=12excos x+sin x+-sin x+cos x dx+Cyex=ex2cos x+sin x+Cy=12cos x+sin x+Ce-xHence, y=12cos x+sin x+Ce-x is the required solution.

Page No 21.106:

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.=eP dx          =e2 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+Cy e2x=152e2x2sin x-cos x+e2x2 cos x+sin x dx+CPutting e2x2 sin x-cos x=t2e2x2sin x-cos x+e2x2 cos x+sin x dx=dty e2x=15dt+Cy e2x=t5+Cy e2x=e2x52sin x-cos x+Cy=152sin x-cos x+Ce-2xHence, y=152sin x-cos x+Ce-2x is the required solution.

Page No 21.106:

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.=eP 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 xcos xdydx-ysin x=-sin 2x Integrating both sides with respect to x, we gety cos x=-sin 2x dx+Cycos x=cos 2x2+C2y cos x=cos 2x+2C2y cos x=cos 2x+K,      where k=2CHence, 2y cos x=cos 2x+K is the required solution.

Page No 21.106:

Question 16:

1+x2dydx+y=tan-1 x

Answer:

We have, 1+x2dydx+y=tan-1xdydx+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.=eP dx          =e11+x2 dx          =etan-1xMultiplying both sides of 1 by etan-1x, we getetan-1x dydx+y1+x2=etan-1x tan-1x1+x2etan-1xdydx+etan-1xy1+x2=etan-1xtan-1x1+x2Integrating both sides with respect to x, we getetan-1xy=tan-1x×etan-1x1+x2 dx+Cetan-1xy=I+C        .....2Here, I=tan-1x×etan-1x1+x2 dxPutting tan-1 x=t, we get11+x2dx=dt I=t et dt     =tetdt-ddttetdtdt     =t et-et     =t-1et     =tan-1x-1etan-1xSubstituting the value of I in 2, we getetan-1xy=tan-1x-1etan-1x+Cy=tan-1x-1+Ce-tan-1xHence, y=tan-1x-1+Ce-tan-1x is the required solution.

Page No 21.106:

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.=eP dx          =etan x dx         = elogsec x=sec xMultiplying both sides of 1 by sec x, we getsec xdydx+y tan x=cos x ×sec xsec xdydx+y sec x tan x=1Integrating both sides with respect to x, we gety sec x=dx+Cy sec x=x+CHence, y sec x=x+C  is the required solution.

Page No 21.106:

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.=eP dx          =ecot x dx         = elogsin x=sin xMultiplying both sides of 1 by sin x, we getsin xdydx+ycot x=sin xx2cot x+2xsin 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+Cy sin x=x2cos xdx-ddxx2cos x dxdx+2xsin x  dx+Cy sin x=x2sin x-2xsin x dx+2xsin x dx+Cy sin x=x2sin x+CHence, y sin x=x2sin x+C is the required solution.

Page No 21.106:

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.=etan x dx =elog sec x=sec xTherefore, solution is given byy×I.F.=x2 cos2 x×I.F. dx+Cy sec x=x2 cos x dx+Cy sec x=I+CWhere, I=x2IIcos x Idx+C I=x2cos x dx-ddxx2cos x dxdx I=x2sin x-2x sin x dx I=x2sin x-2xI sin xII dx I=x2sin x-2xsin x dx+2ddxxsin x dxdx I=x2sin x+2x cos x -2cos 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+Cy sec x=x2sin x+2x cos x-2sin x+C

Page No 21.106:

Question 20:

1+x2dydx+y=etan-1 x

Answer:

We have, 1+x2dydx+y=etan-1xdydx+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.=eP dx          =e11+x2 dx          =etan-1xMultiplying both sides of 1 by etan-1x, we getetan-1x dydx+y1+x2=etan-1xetan-1 x1+x2etan-1xdydx+y etan-1x1+x2=etan-1xetan-1x1+x2Integrating both sides with respect to x, we gety etan-1x=e2tan-1x1+x2 dx+Cy 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+C2y etan-1x=e2tan-1x+2C2y etan-1x=e2tan-1x+k, where k=2CHence, 2y etan-1x=e2tan-1x+k is the required solution.

Page No 21.106:

Question 21:

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

Answer:

We have, x dy=2y+2x4+x2dxdydx=2xy+2x3+xdydx-2xy=2x3+x        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-2xQ=2x3+x I.F.=eP dx          =e-2x dx          =e-2log x         =1x2Multiplying both sides of 1 by 1x2, we get1x2 dydx-2xy=1x2 2x3+x1x2dydx-2x3y=2x+1xIntegrating both sides with respect to x, we get1x2y=2x+1xdx+C1x2y=x2+log x+Cy=x4+x2log x+Cx2Hence, y=x4+x2log x+Cx2 is the required solution.

Page No 21.106:

Question 22:

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

Answer:

We have,1+y2+x-etan-1ydydx=0x-etan-1ydydx=-1+y2dydx=-1+y2x-etan-1ydxdy=-x-etan-1y1+y2dxdy+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.=eP dy          =e11+y2 dy          =etan-1yMultiplying both sides of 1 by etan-1y, we getetan-1y dxdy+x1+y2=etan-1y etan-1 y1+y2etan-1ydxdy+x etan-1x1+y2=e2tan-1 y1+y2Integrating both sides with respect to y, we getx etan-1y=e2tan-1 y1+y2 dy+Cx 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+C2x etan-1y=e2tan-1y+2C2x etan-1y=e2tan-1y+k     where k=2CHence, 2x etan-1y=e2tan-1y+k is the required solution.

Page No 21.106:

Question 23:

y2dxdy+x-1y=0

Answer:

We have,y2dxdy+x-1y=0y2dxdy+x=1y dxdy+1y2x=1y3         .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1y2Q=1y3 I.F.=eP dy          =e1y2dy         = 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+Cx e-1y=I+C         .....2whereI= e-1y1y3dyPutting t=1y, we getdt=-1y2dy I=- t Ie-tIIdt    =-te-tdt+ddtte-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 21.106:

Question 24:

2x-10y3dydx+y=0

Answer:

We have, 2x-10y3dydx+y=02x-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.=eP dy          =e2ydy         = 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+Cxy2=2y5+Cx=2y3+Cy-2Hence, x=2y3+Cy-2 is the required solution.

Page No 21.106:

Question 25:

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

Answer:

We have,x+tan ydy=sin 2y dxdxdy=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.=eP 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+Cxtan y=I+C         .....2where I=12 1tan y×sec2y dyPutting t=tan y, we getdt=sec2 y dy I=121t×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 21.106:

Question 26:

dx + xdy = ey 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.=eP dy          =edy          =eyMultiplying both sides of 1 by ey, we getey dxdy+x=ey e-ysec2yeydxdy+x ey=sec2yIntegrating both sides with respect to y, we getx ey=sec2y dy+Cx ey=tan y+CHence, x ey=tan y+C is the required solution.

Page No 21.106:

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.=eP 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 xcos xdydx+ysin x=-sin 2xIntegrating both sides with respect to x, we gety cos x=-sin 2x dx+Cy cos x=cos 2x2+Cy cos x=1-2sin2x2+Cy cos x=-sin2x+12+Cy cos x=-sin2x+K     where k=12+Cy=sec x-sin2x+KHence, y=sec x-sin2x+K is the required solution.

Page No 21.106:

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.=eP dx          =ecosx 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+Cy esin x=I+C           .....2whereI=esin x sin x cos x dxPutting t=sin x, we getdt=cos x dx I=etII tI dt     =tetdt-ddttetdtdt     =t et-et     =ett-1     = esin xsin x-1Putting the value of I in 2, we gety esin x= esin xsin x-1+Cy=sin x-1+Ce-sin x Hence, y=sin x-1+Ce-sin x is the required solution.

Page No 21.106:

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+1dxy11+x2=1+11+x2dxy11+x2=x+tan-1x+Cy=x+tan-1x+C1+x2

Page No 21.106:

Question 30:

sin xdydx+y cos x=2 sin2 x cos x

Answer:

We have,sin xdydx+y cos x=2 sin2 x cos xdydx+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.=eP dx          =ecot x dx         = elogsin x=sin xMultiplying both sides of 1 by sin x, we getsin xdydx+y cot x=sin x×2sin xcos xsin xdydx+y cosx=2sin2 xcos xIntegrating both sides with respect to x, we gety sin x=2sin2 x cos x dx+C           .....2Putting sin x=tcos x dx=dtTherefore, 2 becomesy sin x=2t2 dt+Cy sin x=23t3+Cy sin x=23sin3x+CHence, y sin x=23sin3x+C is the required solution.

Page No 21.106:

Question 31:

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

Answer:

We have,x2-1dydx+2x+2y=2x+1dydx+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.=eP dx          =e2x+2x2-1 dx          =e2xx2-1+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-1x-13x+1dydx-2x+2x-12x+12y=2x-12x+1Integrating both sides with respect to x, we getx-13x+1y=2x-12x+1 dx+Cx-13x+1y=2x+12-4xx+1 dx+Cx-13x+1y=2x+1-4xx+1 dx+Cx-13x+1y=2x+1-4x+1-1x+1 dx+Cx-13x+1y=2x+1-4+4x+1 dx+Cx-13x+1y=2x-6+8x+1 dx+Cx-13x+1y=x2-6x+8log x+1+Cy=x+1x-13x2-6x+8log x+1+CHence, y=x+1x-13x2-6x+8logx+1 +C is the required solution.

Page No 21.106:

Question 32:

xdydx+2y=x cos x

Answer:

We have, xdydx+2y=x cos xdydx+2xy=cos xComparing with dydx+Py=Q, we getP=2xQ=cos xNow, I.F.=e2xdx =e2log x=x2Solution is given by,y×I.F.=cos x×I.F. dx+Cyx2=x2 cos x dx+Cx2y=I+C          .....1Where,I=x2IIcos x Idx+C I=x2cos x dx-ddxx2cos x dxdx I=x2sin x-2x sin x dx I=x2sin x-2xI sin xII dx I=x2sin x-2xsin x dx+2ddxxsin x dxdx I=x2sin x+2x cos x -2cos x dx I=x2sin x+2x cos x -2sin x I=x2sin x+2x cos x -2sin xTherefore 1 becomesx2y=x2sin x+2x cos x-2sin x+C

Page No 21.106:

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.=eP dx          =e-dx          =e-xMultiplying both sides of 1 by e-x, we gete-xdydx-y=x exe-xe-xdydx-e-xy=xIntegrating both sides with respect to x, we gete-xy=x dx+Ce-xy=x22+Cy=x22+Ce xHence, y=x22+Ce x is the required solution.

Page No 21.106:

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.=eP dx          =e2dx         = 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+Ce2xy=xe6xdx-ddxxe6xdxdx+Ce2xy=xe6x6-e6x36+Cy=xe4x6-e4x36+Ce-2xHence, y=xe4x6-e4x36+Ce-2x is the required solution.

Page No 21.106:

Question 35:

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

Answer:

 We have, x+2y2dydx=ydxdy=1yx+2y2  dxdy-1yx=2y        .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-1yQ=2y I.F.=eP dy          =e-1ydy         = e-log y=1yMultiplying both sides of 1 by 1y, we get1ydxdy-1yx= 1y×2y1ydxdy-1y2x=2Integrating both sides with respect to y, we getx1y= 2dy+Cx1y=2y+Cx=2y2+Cy        .....2Now,y=1 at x=2 2=2+CC=0Putting the value of C in 2, we getx=2y2Hence, x=2y2 is the required solution.



Page No 21.107:

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.=eP dx          =e3 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+30  ye3x=em+3xm+3+Cy=emxm+3+Ce-3x ye3x=e0×x dx+C    when m+3=0 ye3x=dx+Cye3x=x+Cy=x+Ce-3xHence, y=emxm+3+Ce-3x, where m+30 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.=eP 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 dxI=12e-xsin 2x+12e-xsin 2x dxI=12e-xsin 2x-14e-xcos 2x-12×12-e-x×-cos 2x dxI=12e-xsin 2x-14e-xcos 2x-14e-xcos 2x dxI=12e-xsin 2x-14e-xcos 2x-14I       From 354I=12e-xsin 2x-14e-xcos 2x5I=2e-xsin 2x-e-xcos 2xI=e-x52sin 2x-cos 2x           .....(4)From (2) and (4) we getye-x=e-x52sin 2x-cos 2x+Cy=152sin 2x-cos 2x+CexHence, y=152sin 2x-cos 2x+Cex is the required solution.


iii We have, xdydx-y=x+1e-xdydx-1xy=x+1xe-x         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1xQ=x+1xe-x I.F.=eP 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=dt1x+1x2e-xdx=-dtTherefore 2 becomes1xy=-dt+C1xy=-t+C1xy=-1xe-x+Cy=-e-x+CxHence, y=-e-x+Cx is the required solution.


iv We have, xdydx+y=x4dydx+1xy=x3         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1x Q=x3 I.F.=eP dx          =e1x 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+Cxy=x55+Cy=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 xdydx+yx log x=1 xdydx+1x log xy=1 xComparing with dydx+Py=Q, we getP=1x log x Q=1 xNow,I.F.=ePdx=e1x log xdx                         =eloglog x                         =log xSo, the solution is given byy×I.F.=Q×I.F. dx+Cy log x=1 x×log x dx+Cy log x=log x22+Cy=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.=eP 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+211+x2dydx-2xy1+x22=x2+2x2+1Integrating both sides with respect to x, we get11+x2y=x2+2x2+1 dx+C11+x2y=x2+1+1x2+1 dx+C11+x2y=dx+1x2+1 dx+C11+x2y=x+tan-1x +Cy=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.=eP dx          =ecos x dx          =esin xMultiplying both sides of 1 by esin x, we getesin x dydx+y cos x=esin x×esin x cos xesin xdydx+yesin xcos x=e2sin x cos xIntegrating both sides with respect to x, we getesin xy=e2sin x cos x dx+Cesin 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+Cy=esin x2+Ce-sin xHence, y=esin x2+Ce-sin x is the required solution.


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


ix We have,dydxcos2x=tan x-ydydx+1cos2 xy=tan x sec2 xdydx+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.=eP dx          =esec2 x dx          =etan xMultiplying both sides of 1 by etan x, we getetan x dydx+y sec2 x=etan xtan x sec2 xetan xdydx+yetan xsec2 x=etan xtan x sec2 xIntegrating both sides with respect to x, we getetan xy=etan xtan x sec2 x dx+Cetan xy=I+C        .....2Where,I=etan xtan x sec2 x dxPutting t= tan x, we getdt=sec2 x dx I=tI etII dt    =tet dt-ddttet dtdt    =t et-et    =t-1et    =tan x-1etan xPutting the value of I in 2, we getetan xy=tan x-1etan x+Cy=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.=eP dy          =edy          =eyMultiplying both sides of 1 by ey, we geteydxdy+x=eye-ysec2yeydxdy+eyx=sec2yIntegrating both sides with respect to y, we geteyx=sec2y dy+Ceyx=tan y+Cx=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 xdydx+yx log x=2 xdydx+1x log xy=2 xComparing with dydx+Py=Q, we getP=1x log xQ=2 xNow, I.F.=ePdx=e1x log xdx                         =eloglog x                         =log xSo, the solution is given by y×I.F.=Q×I.F. dx+Cy log x=21 x×log x dx+CPutting log x=t1xdx=dty log x=2t dt+Cy log x=2t22+Cy log x=t2+Cy log x=log x2+C    log x=ty=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.=ePdx=e2xdx                         =e2logx                         =x2So, the solution is given byy×I.F.=Q×I.F. dx+Cx2y=x3IIlog xI dx+Cx2y=log xx3 dx-ddxlog xx3 dxdx+Cx2y=x4log x4-x34dx+Cx2y=x4log x4-x416+Cy=x2log x4-x216+Cx2y=x2164log x-1+Cx2

Page No 21.107:

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;tanx0 given that y = 0 when x=π2.

Answer:

i We have,y'+y=exdydx+y=ex           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=1 and Q=ex I.F.=eP dx          =e1 dx         = exMultiplying both sides of 1 by I.F.=ex, we getex dydx+y=exexexdydx+exy=e2xIntegrating both sides with respect to x, we gety ex=e2x dx+Cy ex=e2x2+C           .....2Now, y0=12 12 e0=e02+CC=0Putting the value of C in 2, we getyex=e2x2ex=ex2Hence, y=ex2  is the required solution.


ii We have,xdydx-y=log xdydx-yx=log xx           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=log xx I.F.=eP dx          =e-1x dx          =e-log x          =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1x×logxx1xdydx-1x2y=logxx2Integrating both sides with respect to x, we gety1x=1x2II×logxI dx+Cyx=log x1x2dx-ddxlog x1x2dxdx+Cyx=-log xx+1x2dx+Cyx=-log xx-1x+Cy=-log x-1+Cx           .....2Now, y1=0 0=-0-1+C1C=1Putting the value of C in 2, we gety=-log x-1+xy=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.=eP dx          =e2 dx         = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xe-2xsin xe2x dydx+2y=sin xIntegrating both sides with respect to x, we getye2x=sin x dx+Cye2x=-cos x+C           .....2Now,y0=0 0×e0=-cos 0+CC=1Putting the value of C in 2, we getye2x=-cos x+1ye2x=1-cos xHence, ye2x=1-cos x is the required solution.


iv We have,xdydx-y=x+1e-xdydx-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.=eP 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=dt1x+1x2e-x dx=-dt1xy=-dt+Cyx=-t+Cyx=-e-xx+Cy=-e-x+Cx         .....2Now, y1=0 0=-e-1+CC=e-1Putting the value of C in 2, we gety=-e-x+xe-1y=xe-1-e-xHence, y=xe-1-e-x is the required solution.


v We have,1+y2dx+x-e-tan-1ydy=0x-e-tan-1ydydx=-1+y21+y2dxdy=-x-e-tan-1ydxdy+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.=eP dy          =e11+y2 dy          =etan-1yMultiplying both sides of 1 by I.F.=etan-1y, we getetan-1y dxdy+x1+y2=etan-1ye-tan-1 y1+y2etan-1y dxdy+x1+y2=11+y2Integrating both sides with respect to y, we getetan-1yx=11+y2 dy+Cxetan-1y=tan-1y+C        .....2Now, y0=0 0×e0=0+CC=0Putting the value of C in 2, we getxetan-1y=tan-1y+0xetan-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.=eP dx          =etan x dx         = elogsec x=sec xMultiplying both sides of 1 by I.F.=sec x, we getsec xdydx+ytan x=sec xx2tan x+2xsec xdydx+ytan x=x2tan x sec x+2x sec xIntegrating both sides with respect to x, we getysec x=x2tan xsec x dx+2xII sec xI dx  +Cy sec x=x2tan x sec x dx+2sec xx dx-2ddxsec xx dxdx+Cy sec x=x2tan x sec x dx+x2sec x-x2tan x sec x dx+Cy sec x=x2sec x+Cy=x2+Ccos x      .....2Now,y0=1 1=0+Ccos 0C=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 xdydx+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.=eP dx          =e1xdx         = elog x=xMultiplying both sides of (1) by I.F.=x, we getxdydx+1xy=xcos x+sin xxxdydx+1xy=x cos x+sin xIntegrating both sides with respect to x, we getxy=x cos x dx+sin x dx+Cxy=x sin x-1sin xdx-cos x+Cxy=x sin x+cos x-cos x+Cxy=x sin x+C          .....2Now, yπ2=1 1×π2=π2sinπ2+CC=0Putting the value of C in 2, we getxy=x sin xy=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.=eP dx          =ecot x dx         = elogsin x=sin xMultiplying both sides of 1 by I.F.=sin x, we getsin xdydx+y cot x=sin x4x cosec xsin xdydx+y cot x=4xIntegrating both sides with respect to x, we gety sin x=4x dx+Cy sin x=2x2+C           .....2Now,yπ2=0 0×sinπ2=2π22+CC=-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.=eP dx          =e2tan x dx         = e2logsec x=sec2xMultiplying both sides of (1) by I.F.=sec2 x, we getsec2 x dydx+2y tan x=sec2 x×sin xsec2 x dydx+2y tan x=tan x sec xIntegrating both sides with respect to x, we gety sec2 x=tan x sec x dx+Cy sec2 x=sec x+C           .....2Now, yπ3=0 0secπ32=secπ3+CC=-2Putting the value of C in 2, we gety sec2 x=sec x-2y=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.=eP dx          =e-3cot x dx         = e-3logsin x=cosec3xMultiplying both sides of 1 by I.F.=cosec3x, we getcosec3xdydx-3y cot x=sin 2xcosec3xcosec3xdydx-3y cot x=2cot x cosec xIntegrating both sides with respect to x, we gety cosec3x=2cot x cosec x dx+C ycosec3x=-2cosec x+Cy=-2sin2x+Csin3x           .....2Now, yπ2=2 2=-2sin2π2+Csin3 π2C=4Putting the value of C in 2, we gety=-2sin2x+4sin3xy=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.=eP dx          =ecotx dx         = elogsinx         =sinxMultiplying both sides of 1 by I.F.=sinx, we getsinxdydx+ycotx=2sinxcosxsinxdydx+ycosx=sin2xIntegrating both sides with respect to x, we getysinx=sin2x dx+Cysinx=-cos2x2+C           .....2Now, yπ2=0  0×sinπ2=-cosπ2+CC=-12Putting the value of C in 2, we getysinx=-cos2x2-122ysinx=-1+cos2x2ysinx=-2cos2xy=-cotxcosxHence, y=-cotxcosx  is the required solution.


xiidy=cosx2-ycosecxdxdydx=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.=eP dx          =ecotx dx         = elogsinx         =sinxMultiplying both sides of 1 by I.F.=sinx, we getsinxdydx+ycotx=2sinxcosxsinxdydx+ycosx=sin2xIntegrating both sides with respect to x, we getysinx=sin2x dx+Cysinx=-cos2x2+C     Hence, ysinx=-cos2x2+C  is the required solution.

(xiii) 
tanxdydx=2xtanx+x2-ydydx+1tanxy=2xtanx+x2tanxdydx+cotxy=2x+x2cotx
This is a linear differential equation of the form dydx+Py=Q.

Integrating factor, I.F. = ePdx=ecotxdx=elogsinx=sinx

The solution of the given differential equation is given by

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

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

y=x2-π24cosecx

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

Page No 21.107:

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.=eP dx          =e2x 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+Cx2y=x44+Cy=x24+Cx-2Hence, y=x24+Cx-2  is the required solution.

Page No 21.107:

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.=eP 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+Cye-x=I+C           .....2Here,I=e-xcos x dx           .....3I=e-xsin x--e-xsin x dxI=e-xsin x+e-xsin x dxI=e-xsin x-e-xcos x--e-x×-cos x dxI=e-xsin x-e-xcos x-e-xcos x dxI=e-xsin x-e-xcos x-I   From  3 2I=e-xsin x-cos xI=e-x2sin x-cos x           .....4From   2 and 4 we getye-x=e-x2sin x-cos x+Cy=12sin x-cos x+CexHence, y=12sin x-cos x+Cex  is the required solution.

Page No 21.107:

Question 40:

Solve the differential equation y+3x2dxdy=x

Answer:

We have, y+3x2dxdy=xdydx=y+3x2xdydx-1xy=3x        .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=3x I.F.=eP 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=3dx+Cyx=3x+CHence, yx=3x+C  is the required solution.



Page No 21.108:

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.=eP dy          =ecot y dy         = elogsin y=sin yMultiplying both sides of 1 by I.F.=sin y, we getsin ydxdy+x cot y=sin yy2cot y+2ysin 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+Cx sin y=y2cos ydy-ddyy2cos y dydy+2y sin y  dy+Cx sin y=y2sin y-2y sin y dy+2ysin y dy+Cx sin y=y2sin y+CNow, y=π2 at x=0 0×sin π2=π42sin π2+CC=-π42Putting the value of C, we getx sin y=y2sin y-π42Hence, x sin y=y2sin y-π42 is the required solution.

Page No 21.108:

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+Cxecot-1y=-ett-1+Cxecot-1y=ecot-1y1-cot-1y+C



Page No 21.134:

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=4πr2dSdt=8π r drdt         .....1Given:dSdtα tdSdt=kt, where k is any constantPutting dSdt=kt in (1), we getkt=8π r drdtkt dt=8πr drIntegrating both sides, we get kt dt=8πr drkt22=8π ×r22+C        .....(2)At t=0 s, r=1 unit and at t=3 s, r=2 units                   Given 0=8π×12+CC=-4πAnd92k=8π×2+C92k=12 πk=83πSubstituting the values of C and k in (2), we get 8t26π=8π ×r22-4π4t23=4r2-4t23=r2-1r2=1+t23r=1+13t2

Page No 21.134:

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=5P100dPdt=0.05P

dPP=0.05dt Integrating both sides with respect to t, we getdPP=0.05dt log P=0.05t +CNow,P=P0 at t=0  log P0=0+CC=log P0Putting the value of C, we getlog P=0.05t+log P0logPP0=0.05tTo find the time when the population will double, we haveP=2P0 log2P0P0=0.05tlog 2=0.05tt=log 20.05=20 log 2 years

Page No 21.134:

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=aPdPP=adtlogP=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+logNlogPN=at         .....2According to the question,log2NN=25aa=125log2=125×0.6931=0.0277Putting a=0.0277 in 2, we getlogPN=0.0277t           ...3For P=500000 and N=100000:log500000100000=0.0277tt=log 50.0277=1.6090.0277=58.08 years=Approximately 58 years

Page No 21.134:

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 NdNdt=λN1NdN=λdtIntegrating both sides, we get1NdN=λdtlog N=λt+log C             .....(1)Initially when t=0, then N=100000 Givenlog 100000=0+log Clog 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 10000012log 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 100000log 2=T2log 1110T=2log 2log1110 The count will reach 200000 in 2log 2log1110 hours.

Page No 21.134:

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=6P100dPdt=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+CC=log P0Putting the value of C, we getlog P=0.06t +log P0logPP0=0.06te0.06t=PP0To find the amount after 10 years, we gete0.06×10=PP0e0.6=PP01.822=PP0P=1.822P0P=1.822×1000=Rs 1822To find the time after which the amount will double, we haveP=2P0 log2P0P0=0.06tlog 2=0.06tt=0.69310.06=11.55 years

Page No 21.134:

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=aPdPP=adtlogP=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+logNlogPN=at         .....2According to the question,log3NN=5aa=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=PNe2.1972=PNPN=9P=9NTo find the time taken when the number of bacteria becomes 10 times of the number of initial population, we haveP=10N log10NN=15tlog 3t=5 log 10log 3

Page No 21.134:

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=βPdPP=βdtlogP=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=10ββ=110log54Putting β=110log 54 in 2, we getlog 200000=1990×110log54+log Clog 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 log54logP=201 log 54+log 200000-199log54logP=log 54201- log54199+log 200000logP=log 5420145199+log 200000logP=log 542+log 200000logP=log2516×200000logP=log  312500P=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+0.15xdC=2+0.15xdxIntegrating both sides with respect to x, we getC=2x+0.152x2+K          .....1At C0=100, we have100=20+0.15202+KK=100Putting the value of T in 1, we getC=2x+0.152x2+100C=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=8P100dPdt=2P25

dPP=225dtIntegrating both sides with respect to t, we getlog P=225t +C          .....1Now,P=P0 at t=0  log P0=0+CC=log P0Putting the value of C in 1, we getlog P=225t +log P0logPP0=225te225t=PP0To find the amount after 1 year, we havee225=PP0e0.08=PP01.0833=PP0P=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=Edidt+RLi=EL            .....1 I.F.=eRL dt          =eRLtMultiplying both sides of (1) by I.F.=eRLt, we get eRLt didt+RLi=eRLt×ELeRLtdidt+eRLtRLi=eRLt×ELIntegrating both sides with respect to t, we geteRLti=ELeRLt dt+CeRLti=EL×LReRLt +CeRLti=EReRLt +C            .....2Now, i=0 at t=0 e0×0=ERe0 +CC=-ERPutting the value of C in 2, we geteRLti=EReRLt -ERi=ER -ERe-RLti=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>0dPP=-adtIntegrating both sides, we getlogP=-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+logNlogNP=atAccording to the question, logNN2=atlog2=att=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=-aPdPP=-adtIntegrating both sides, we getlogP=-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+logNlogPN=-at         .....2According to the question,P=12N at t=1590logN2N=-1590a-log 2=-1590aa=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=0.9996P=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=-xyy dy=-x dx Integrating both sides with respect to x, we gety dy=-x dxy22=-x22+CSince the curve passes through 3,-4, it satisfies the above equation. -422=-322+C8=-92+CC=252Putting the value of C, we gety22=-x22+252x2+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-ydyy1-y=dxx+1

Integrating both sides, we getdyy1-y=dxx+11y+11-ydy=dxx+1log 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+CC=log 23Putting the value of C in 1, we getlog y-log 1-y=log x+1+log 23log y1-y=log 2x+13y1-y=2x+13y1-y=±2x+13y1-y=2x+13  or  y1-y=-2x+13Here, given point 2, 2 does not satisfy y1-y=2x+13But it satisfy y1-y=-2x+13y1-y=-2x+13yy-1=2x+133y=2x+1y-13y=2xy-2x+2y-22xy-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-cos2yxdydx=yx-cos2yx

Let y=vxdydx=v+xdvdx v+xdvdx=v-cos2vxdvdx=-cos2vsec2 v dv=-1xdxIntegrating both sides with respect to x, we getsec2 v dv=-1xdxtan v=-log x+Ctan yx=-log x+CSince the curve passes through 1, π4, it satisfies the above equation. tan π4=-log 1+CC=1Putting the value of C, we gettan yx=-log x+1tan yx=-log x+log etan 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=-ydxdyX=x-ydxdyTherefore, cut off by the tangent on the x-axis = x-ydxdyGiven, x-ydxdy=4yTherefore, -ydxdy=4y - xdxdy=x-4yydydx=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=v1-4vxdvdx=v1-4v-v = 4v21-4v1-4vv2dv=4dxx
Integrating on both sides we get,
1-4vv2dv=4dxxTherefore, dvv2-4dvv=4dxx-1v-4 log v = 4 logx + log c-1v = 4 logx + log c +4 log v4 log(xv) + log c = -1vputting the value of v we get4 log(x×yx) + log c = -xy4 log(y) + log c = -xylog (y4c) = -xyy4c = 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+2xdydx-y=2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1 and Q=2x I.F.=eP 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=2e-xIIxI  dx+Cye-x=2xe-xdx-2ddxxe-xdxdx+Cye-x=-2xe-x-2e-x+C           .....2Since the curve passes through origin, we have0×e0=-2×0×e0-2e0+CC=2Putting the value of C in 2, we getye-x=-2xe-x-2e-x+2y=-2x-2+2exy+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+3ydydx=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.=eP dx          =e-3 dx         = e-3xMultiplying both sides of (1), by I.F.=e-3x, we gete-3x dydx-3y=e-3x.2xe-3x dydx-3y=2xe-3xIntegrating both sides with respect to x, we gety e-3x=2xe-3x dx+Cy e-3x=2xIe-3xII dx+Cy e-3x=2xe-3x dx-2ddxxe-3x dxdx+Cy e-3x=-2xe-3x3+2×13e-3x dx+Cy e-3x=-23xe-3x-2×19e-3x+Cy e-3x=-23xe-3x-29e-3x+CSince the curve passes through 1, 2, it satisfies the above equation. 2e-3=-23e-3-29e-3+CC=2e-3+23e-3+29e-3C=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 =xydxdy=OT

It is given, OT = 2x

xydxdy=2xx=ydxdy-dxx=dyyxy=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+1dydx-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 +Cx+1xy=x+1x dx +Cx+1xy=dx+1xdx +Cx+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+CC=-1Putting the value of C in the equation of the curve, we getx+1xy=x+log x-1y=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 get12ydy=1xdx12log y=log x+CSince the curve passes through 3,-4, it satisfies the above equation. 1 2log -4=log 3+Clog 2-log 3=CC=log 23Putting the value of C, we getlog y=2log x+2log 23log y=log 49x2y=±49x29y-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 +Cye-3x=x-1e-3x dx +Cye-3x =xIe-3xIIdx-e-3x dx+Cye-3x=xe-3x dx-ddxxe-3x dxdx-e-3x dx+Cye-3x=-13xe-3x+13e-3x dx-e-3x dx+Cye-3x=-13xe-3x-19e-3x +13e-3x+Cy=-13x-19 +13+Ce3xy=-13x +29+Ce3xSince the curve passes throught the origin, it satisfies the equation of the curve.0=-0+29+Ce0C=-29Putting the value of C in the equation of the curve, we gety=-13x +291-e3xy+13x=291-e3x33y+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+xydydx=x1+y

11+ydy=x dxIntegrating both sides with respect to x, we get11+ydy=x dxlog 1+y=x22+CSince the curve passes through 0, 1, it satisfies the above equation. log 1+1=02+CC=log 2Putting the value of C, we getlog 1+y=x22+log 2log 1+y2=x221+y2=ex22y+1=2ex22

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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 dydx2y2-2xydydx-x2 =0          Hence proved.Now, y2-2xydydx-x2 =0    dydx = y2-x22xy2xydydx-y2 =-x2 2ydydx-y2 x=-x Let y2=vdvdx-vx=-x       
Multiplying by the integrating factor e-1xdx=1x
v.1x=-x.1xdx + c= -x+cy2x2=-x+cx2 + y2=cx

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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

CoordinateofT=xydxdy, 0AT=xxydxdy=2xEquationoftangent,Yy=dydx(Xx)y-0=dydxxxydxdyydxdy=2xdxx=2dyylnx=lny2+lncx=cy2As the circle passes through 1, 2.1=c×22c=144x=y2

 

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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-xydydx=3-xy dy=3-xdxy22=3x-x22+C          .....1Since the curve passes through the point 3, 4, it satisfies the equation.422=33-322+CC=8-9+92C=92-1=72Putting the value of C in 1, we gety22=3x-x22+72y2=6x-x2+7x2+y2-6x-7=0

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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=aPdPP=adtlog 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 Nlog PN=at         .....2According to the question,log 2NN=6aa=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 2log PN=3log 2log PN=log 8PN=8P=8N

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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=-aPdPP=-a dtIntegrating both sides, we getlog 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 Nlog PN=-at         .....2According to the question,P=98.9100N=0.989N at t=25log 0.989NN=-25aa=-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=12Plog NN2=125log 0.989tlog 2=125log 0.989t  t=25log 2log 0.989=25×0.69310.01106=1566.681567 approx.

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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+v2x22vx2xdvdx=1+v22v-vx dvdx=1+v2-2v22vxdvdx=1-v22v2v1-v2dv=1xdx

Integrating both sides, we get2v1-v2dy=1xdx-log 1-v2=log x-log Clog 1-v2C=-log x1-v2=Cx1-yx2=Cxx2-y2x2=Cxx2-y2=Cx
Thus, x2-y2=Cx is the equation of the rectangular hyperbola.



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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 +Cye-x=xIe-xIIdx+Cye-x=xe-x dx-ddxxe-x dxdx+Cye-x=-xe-x+e-x dx+Cye-x=-xe-x-e-x+CSince the curve passes throught the origin, it satisfies the equation of the curve.0e0=-0e0-e0+CC=1Putting the value of C in the equation of the curve, we getye-x=-xe-x-e-x+1ye-x+xe-x+e-x=1y+x+1e-x=1x+y+1=ex

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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 +Cye-x22=xe-x22dx+Cye-x22=I+CNow,I=xe-x22dxPutting -x22=t, we get-xdx=dt I=-etdtI=-etI=-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+CC=2Putting the value of C in the equation of the curve, we getye-x22=-e-x22+2y=-1+2ex22

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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 getdy=x2dxy=x33+CSince the curve passes through -1, 1, it satisfies the above equation. 1=-13+CC=1+13C=43Putting the value of C, we gety=x33+433y=x3+4

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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=xy dy=x dxIntegrating both sides with respect to x, we gety dy=x dxy22=x22+CSince the curve passes through 0, a, it satisfies the above equation. a22=02+CC=a22Putting the value of C, we gety22=x22+a22x2-y2=-a2

Page No 21.136:

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 = dxlog y = x + log Clog y = log ex + log Cy = Cex
Since, the curve passes through (1,1). Therefore, x=1 and y=1 . 
Putting these values in equation obtained above we get,
1=Ce1C=1eputting these values in the equation we get,y=ex-1



Page No 21.137:

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 21.138:

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-3xdydx=-3e-3xd2ydx2=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 21.138:

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'=exy''=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 21.138:

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=mdydx=yx               From 1xdydx=yxdydx-y=0

Page No 21.138:

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+bd2ydx2=-a sinx+b    d2ydx2=-a×ya          Using 2d2ydx2=-y d2ydx2+y=0

Page No 21.138:

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×xy2=2xydydxy2-2xydydx=0

Page No 21.138:

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=0x+y-adydx=0x=a-ydydxxdydx=a-ya=y+xdydx 
Substituting the value of a in (1), we get

x2+y-y-xdydx2=32x2+x2dydx2=9x2dydx2+x2=9dydx2x2dydx2-9dydx2+x2=0x2-9dydx2+x2=0x2-9y'2+x2=0Hence, x2-9y'2+x2=0  is the required differential equation.

Page No 21.138:

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'×yxy'=2yxy'-2y=0

Page No 21.138:

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'=0xa2+yb2y'=0                          .....21a2+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 21.138:

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=01a2=1b2yy''+y'2b2a2=yy''+y'2                            .....(3)
From (2), we get
2xa2=2yb2y'b2a2=yxy'                                    .....(4)
From (3) and (4), we get
yxy'=yy''+y'2yy'=xyy''+xy'2Hence, xyy''+xy'2-yy'=0 is the required differential equation.

Page No 21.138:

Question 11:

Verify that xy = a ex + b ex + 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 getxdydx+y=aex-be-x+2xAgain differentiating with respect to x on both sides, we getxd2ydx2+dydx+dydx=aex+be-x+2xd2ydx2+2dydx=xy-x2+2       xy=aex+be-x+x2xd2ydx2+2dydx-xy+x2-2=0
Thus, xy = a ex + b ex + x2 is the solution of the given differential equation.

Page No 21.138:

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 21.138:

Question 13:

Show that y2x2xy = a is a solution of the differential equation x-2ydydx+2x+y=0.

Answer:

We have,
x-2ydydx+2x+y=0
Now,y2x2xy = a
2ydydx-2x-y-xdydx=02y-xdydx-2x-y=02y-xdydx=2x+yx-2ydydx=-2x+yx-2ydydx+2x+y=0
 Thus, y2x2xy = a is the solution of the given differential equation.

Page No 21.138:

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 21.138:

Question 15:

Find the differential equation corresponding to y = ae2x + be3x + cex where a, b, c are arbitrary constants.

Answer:

We have,
y
= ae2x + be3x + cex            .....(1)
Differentiating with respect to x, we get
dydx=2ae2x-3be-3x+cex         .....2d2ydx2=4ae2x+9be-3x+cexd3ydx3=8ae2x-27be-3x+cexd3ydx3=72ae2x-3be-3x+cex-6ae2x+be-3x+cexd3ydx3=7dydx-6y      Using 1 and 2d3ydx3-7dydx+6y=0

Page No 21.138:

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=4adydx1=2ad2ydx20=2ad3ydx3d3ydx3=0                                  

Page No 21.138:

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 21.139:

Question 18:

dydx=sin3 x cos4 x+xx+1

Answer:

We have,dydx=sin3 x cos4 x+xx+1dy=sin3 x cos4 x+xx+1dxIntegrating both sides, we getdy=sin3 x cos4 x+xx+1dxy=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=dxI2=2t2-1t2dt     =2t4-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 21.139:

Question 19:

dydx=1x2+4x+5

Answer:

We have,dydx=1x2+4x+5dydx=1x2+4x+4+1dydx=1x+22+12dy=1x+22+12dxIntegrating both sides, we gety=1x+22+12dxy=tan-1 x+21+Cy=tan-1 x+2+C

Page No 21.139:

Question 20:

dydx=y2+2y+2

Answer:

We have,dydx=y2+2y+2dydx=y2+2y+1+1dydx=y+12+121y+12+12dy=dxIntegrating both sides, we get1y+12+12dy=dxtan-1 y+11+C=xx=tan-1y+1+C

Page No 21.139:

Question 21:

dydx+4x=ex

Answer:

We have,dydx+4x=exdydx=ex-4xdy=ex-4xdxIntegrating both sides, we getdy=ex-4xdxy=ex-2x2+Cy+2x2=ex+C

Page No 21.139:

Question 22:

dydx=x2 ex

Answer:

We have,dydx=x2exdy=x2ex dxIntegrating both sides, we getdy=x2IexII dxdy=x2ex dx-ddxx2ex dxdxy=x2ex-2xex dxy=x2ex-2xI exII dxy=x2ex-2xex dx+2ddxxex dxdxy=x2ex-2xex+2ex+Cy=x2 -2x+2ex+C

Page No 21.139:

Question 23:

dydx-x sin2 x=1x log x

Answer:

We have, dydx-x sin2 x=1x log xdydx=1x log x+x sin2 xdydx=1x log x+x 21-cos 2xdydx=1x log x+x 2-x 2cos 2xIntegrating both sides, we getdy=1x log x+x 2-x 2cos 2x dxdy=1x log xdx+12x dx-12x cos 2xdxdy=1x log xdx+12x dx-12xI×cos 2xII dx y=log log x+x24-x2cos 2xdx+12ddxxcos 2x dxdxy=log log x+x24-x sin 2x4-cos 2x8+C

Page No 21.139:

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 xdy=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=t2 tan x sec2x+ 2sec2x dx=dt21+tan xsec2 x dx=dtTherefore 1 becomes, dy=1t dty=log t+Cy=log tan2 x+2 tan x+5+C

Page No 21.139:

Question 25:

dydx=sin3 x cos2 x+xex

Answer:

We have,dydx=sin3 x cos2 x+xexdy=sin3 x cos2 x+xexdxIntegrating both sides, we getdy=sin3 x cos2 x+xexdxy=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 dxI1=-t2 1-t2dt       =-t2+t4dt       =-t33+t55+C1       =cos5 x5-cos3 x3+C1 I2=xexdx    =xI exII dx    =xex dx-ddxxex dxdx    =xex-ex+C2    =x-1ex+C2Putting the value of I1 and I2 in 1, we gety=cos5 x5-cos3 x3+C1+x-1ex+C2y=cos5 x5-cos3 x3+x-1ex+C, where C=C1+C2

Page No 21.139:

Question 26:

tan y dx + tan x dy = 0

Answer:

We have,
tan y dx + tan x dy = 0
tan xdydx=-tan y cot y dy=-cot x dxIntegrating both sides, we getcot y dy=-cot x dxlog sin y=- log sin x+log Clog sin y+ log sin x=log Clog sin ysin x=log Csin ysin x=Csin x sin y=C

Page No 21.139:

Question 27:

(1 + x) y dx + (1 + y) x dy = 0

Answer:

We have, 
(1 + x) y dx + (1 + y) x dy = 0

dydx=-y1+xx1+y1+yydy=-1+xxdx1y+ydy=-1x+1dxIntegrating both sides, we get1y+1dy=-1x+1dx1ydy+dy=-1xdx-dxlogy+y=-logx-x+Clogxy+y+x=Cx+y+logxy=C

Page No 21.139:

Question 28:

x cos2 y dx = y cos2 x dy

Answer:

We have,
x cos2 y dx = y cos2 x dy
y sec2y dy=x sec2x dxIntegrating both sides, we getyI sec2yIIdy=xI sec2xII dxysec2ydy-dydy×sec2y dydy=xsec2x dx-dxdx×sec2x dxdxy tan y-tan y dy=x tan x-tan x dx-Cy tan y-log sec y=x tan x-log sec x-Cx tan x-y tan y=logsec x-logsec y+C

Page No 21.139:

Question 29:

cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy

Answer:

We have,
cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy
logsec y+tan ycos ydy=logsec x+tan xcos xdxIntegrating both sides, we getlogsec y+tan ycos ydy=logsec x+tan xcos xdx     .....1Putting logsec y+tan y=t and logsec x+tan x=usec2y+sec y tan ysec y+tan ydy=dt and sec2x+sec x tan xsec x+tan xdx=dusec y dy=dt and sec x dx=duTherefore, 1 becomest dt=u dut22=u22+Clogsec y+tan y22=logsec x+tan x22+Clogsec y+tan y2=logsec x+tan x2+2Clogsec y+tan y2=logsec x+tan x2+k, where k=2C

Page No 21.139:

Question 30:

cosec x (log y) dy + x2y dx = 0

Answer:

We have,
cosec x log ydy+x2y dx=0log yydy=-x2 cosec xdxlog yydy=-x2 sin x dxIntegrating both sides, we getlog yydy=-x2 sin x dx      .....1Putting log y=t1ydy=dtTherefore,  1 becomest dt=-x2 sin x dxt22=-x2I sin xII dx12log y2 =-x2 sin x dx-ddxx2sin x dxdx12log y2 =x2 cos x+2x cos x dx12log y2 =x2cos x-2xI cos xII dx12log y2 =x2cos x-2xcos x dx+2ddxxcos x dxdx12log y2 =x2cos x-2x sin x-2cos x+C12log y2 =x2-2cos x-2x sin x+C12log y2+2-x2cos x+2x sin x=C

Page No 21.139:

Question 31:

(1 − x2) dy + xy dx = xy2 dx

Answer:

We have,
1-x2dy+xy dx=xy2dx1-x2dy=xy2dx-xy dx1-x2dy=xy2-ydx1y2-ydy=x1-x2dxIntegrating both sides, we get1y2-ydy=x1-x2dx1y2-y+14-14dy=x1-x2dx1y-122-122dy=-12-2x1-x2dx12×12log y-12-12y-12+12=-12log 1-x2+log Clog y-1y=-12log 1-x2+log C2 log y-1y=-log 1-x2+2 log Clog y-12y2=-log1-x2+2 log Clog y-12y2+log 1-x2=log C2logy-121-x2y2=log C2y-121-x2y2=C2y-121-x2=y2C2

Page No 21.139:

Question 32:

dydx=sin x+x cos xy2 log y+1

Answer:

We have,
dydx=sin x+x cos xy2 log y+1y 2 log y+1dy= sin x+x cos xdxIntegrating both sides, we get2yII log y Idy+y dy=sin x dx+x cos x dx2log yy dy-2ddylog y×y dydy+y dy=-cos x+x cos x dx-dxdx×cos x dxy2log y-y dy+y dy=-cos x+x sin x dx+cos x+Cy2log y=x sin x+C

Page No 21.139:

Question 33:

x (e2y − 1) dy + (x2 − 1) ey dx = 0

Answer:

We have,xe2y-1dy+x2-1eydx=0xe2y-1dy=1-x2eydxe2y-1eydy=1-x2xdxey-e-ydy=1x-xdxIntegrating both sides, we getey-e-ydy=1x-xdxey+e-y=log x-12x2+Cey+e-y-log x+12x2=C

Page No 21.139:

Question 34:

dydx+1=ex+y

Answer:

We have,
dydx+1=ex+y     .....1Let x+y=v1+dydx=dvdxdydx=dvdx-1Then, 1 becomesdvdx-1+1=evdvdx=eve-vdv=dxIntegrating both sides, we gete-vdv=dx-e-v=x+C-1=evx+C-1=x+Cex+y

Page No 21.139:

Question 35:

dydx=x+y2

Answer:

We have,dydx=x+y2          .....1Let x+y=v1+dydx=dvdxdydx=dvdx-1Therefore, 1 becomes dvdx-1=v2dvdx=v2+11v2+1dv=dxIntegrating both sides, we get1v2+1dv=dxtan-1 v=x+Cv=tanx+Cx+y=tanx+C     

Page No 21.139:

Question 36:

cos (x + y) dy = dx

Answer:

We have,cos x+ydy=dxdydx=1cosx+y     .....1Let x+y=v1+dydx=dvdxdydx=dvdx-1Therefore,  1 becomes dvdx-1=1cos vdvdx=cos v+1cos vcos vcos v+1dv=dxIntegrating both sides, we getcos vcos v+1dv=dxcos v1-cos v1-cos2 vdv=dxcos v1-cos vsin2 vdv=dxcot v cosec v-cot2vdv=dxcot v cosec v-cosec2 v+1dv=dx-cosec v+cot v+v=x+C-cosec x+y+cot x+y+x+y=x+C-cosec x+y+cot x+y+y=Ccosec x+y-cot x+y=y-C1-cos x+ysin x+y=y-C2 sin2 x+y22 sin x+y2 cos x+y2=y-Csin x+y2cos x+y=y-Ctanx+y2=y-C

Page No 21.139:

Question 37:

dydx+yx=y2x2

Answer:

We have,dydx+yx=y2x2dydx=yx2-yxPutting y=vx, we getdydx=v+xdvdx v+xdvdx=v2-vxdvdx=v2-2v1v2-2v dv=1xdxIntegrating both sides, we get1v2-2v dv=1xdx1v2-2v+1-1 dv=1xdx1v-12-12 dv=1xdx12log v-1-1v-1+1 =log x+log Clog v-2v12 =log Cxlog yx-2yx12 =log Cxlog y-2xy12 =log Cxy-2xy12=Cxy-2xy =C2x2y-2x=kx2y, where k=C2

Page No 21.139:

Question 38:

dydx=yx-yxx+y

Answer:

We have,dydx=yx-yxx+yPutting y=vx, we getdydx=v+xdvdx v+xdvdx=vxx-vxxx+vxv+xdvdx=v1-v1+vxdvdx=v1-v1+v-vxdvdx=v-v2-v-v21+vxdvdx=-2v21+v-1+v2v2 dv=1xdxIntegrating both sides, we get-1+v2v2 dv=1xdx-121v2dv-121vdv=1xdx12v-12log v =log x+log Cxy-log yx =2log x+2log Cxy-log yx =log C2x2xy =log C2x2+log yxxy =log C2xyexy =C2xyexy =k xy, where k=C2

Page No 21.139:

Question 39:

(x + y − 1) dy = (x + y) dx

Answer:

We have,x+y-1dy=x+ydxdydx=x+yx+y-1Putting x+y=v, we get1+dydx=dvdxdydx=dvdx-1 dvdx-1=vv-1dvdx=vv-1+1dvdx=v+v-1v-1dvdx=2v-1v-1v-12v-1 dv=dxIntegrating both sides, we getv-12v+1 dv=dx122v2v-1dv-12v-1dv=dx122v-1+12v-1dv-12v-1dv=dx12dv+1212v-1dv-12v-1dv=dx12dv-1212v-1dv=dx12v-14log 2v-1 =x+C12x+y-14log 2x+2y-1 =x+C2x+y-log 2x+2y-1 =4x+4C2x+y-4x-log 2x+2y-1 =4C      2y-x-log 2x+2y-1 =k, where k=4C

Page No 21.139:

Question 40:

dydx-y cot x=cosec x

Answer:

We have,dydx-y cot x=cosec xComparing with dydx+Py=Q, we getP=-cot x Q=cosec xNow,I.F.=e-cot x dx=e-log sin x=elog cosec x=cosec xSo, the solution is given byy cosec x=cosec x×cosec x dx+Cy cosec x=cosec2 x dx+Cy cosec x=-cot x +C

Page No 21.139:

Question 41:

dydx-y tan x=-2 sin x

Answer:

We have,dydx-y tan x=-2sin xComparing with dydx+Py=Q, we getP=-tan x Q=-2sin xNow,I.F.=e-tan xdx=e-logsec x=elogcos x=cos xSo, the solution is given byy cosx=-2sin x cos x dx+Cy cosx=-sin 2x dx+C y cosx=cos 2x2+C

Page No 21.139:

Question 42:

dydx-y tan x=ex sec x

Answer:

We have,dydx-y tan x=ex sec xComparing with dydx+Py=Q, we getP=-tan x Q=ex sec xNow,I.F.=e-tan xdx=e-logsec x=elogcos x=cos xSo, the solution is given by y cosx=cos x exsec x dx+Cy cosx=ex dx+C y cosx=ex+C

Page No 21.139:

Question 43:

dydx-y tan x=ex

Answer:

We have,dydx-y tan x=exComparing with dydx+Py=Q, we getP=-tan x Q=exNow,I.F.=e-tan x dx=e-logsec x=elogcos x=cos xSo, the solution is given byy cos x=excos x dx+Cy cosx=I+C           .....1Where,I=exIIcos xI dx            .....2 I=cos xex dx-ddxcos xex dxdx I=cos x ex+sin x ex dx I=cos x ex+sin xI exII dx I=cos x ex+sin xex dx-ddxsin xex dxdx I=cos x ex+sin xex -cos x ex dx I=cos x ex+sin xex -I        From 2 2I=cos x ex+sin xex  I=ex 2cos x+sin x  y cos x=ex 2cos x+sin x +C        From 1

Page No 21.139:

Question 44:

(1 + y + x2 y) dx + (x + x3) dy = 0

Answer:

1+y+x2ydx+x+x3dy=0dx+y1+x2dx+x1+x2dy=0dx+1+x2 ydx+xdy=01+x2 ydx+xdy=-dxydx+xdy=-11+x2dxydx+xdy=-dx1+x2

On integrating both side we get,

xy=-11+x2dxxy=-tan-1x+cxy+tan-1x=c

Page No 21.139:

Question 45:

(x2 + 1) dy + (2y − 1) dx = 0

Answer:

We have,1+x2dy+2y-1dx=01+x2dy=1-2ydxdy1-2y=11+x2dxIntegrating both sides, we get11-2ydy=11+x2dx-12log1-2y=tan-1x-log C-log1-2y=2tan-1x-2log C-2tan-1x=-log C+log1-2y-2tan-1x=log 1-2yCe-2tan-1x=1-2yCCe-2tan-1x=1-2y1-Ce-2tan-1x=2y12-C2e-2tan-1x=yy=12+Ke-2tan-1x, where K=-C2

Page No 21.139:

Question 46:

y sec2 x + (y + 7) tan x dydx = 0

Answer:

We have,y sec2x+y+7tan xdydx=0y sec2x=-y+7tan xdydx-y-7ydy=sec2xtan xdx-1-7ydy=sec2xtan xdxIntegrating both sides, we get-1-7ydy=sec2xtan xdx-y-7log y=log tan x+log C-y=log tan x+logy7+log C-y=logCy7tan xe-y=Cy7tan xy7tan x=e-yCy7tan x=ke-y, where k=1C

Page No 21.139:

Question 47:

(2ax + x2) dydx = a2 + 2ax

Answer:

2ax+x2dydx=a2+2axdydx=a2+2ax2ax+x2=aa+2xx2a+xLetx=2atan2θ dx=4a tanθsec2θ dθdydx=aa+4atan2θ2a tan2θ 2a1+tan2θdy=a 1+4 tan2θ2 tan2θ 2a sec2θdxdy=a1+4 tan2θ2 tan2θ 2a sec2θ4atanθ sec2θdθ=a 1+4tan2θtanθdθ=a1tanθ+4tanθdθy=acotθ+4 tanθ dθy=alog sinθ+4 log cosθ+cy=alogsinθ4log cosθ+cAs,x=2a tan2θtanθ=x2ay=a log sinθcos4θ+c=alogtanθcos3θ+c=alog x2a×x+2a2a3+cy=alogx12(x+2a)324a2+cy+C=a2 log x+3logx+2a          where C=c-alog4a2

Page No 21.139:

Question 48:

(x3 − 2y3) dx + 3x2 y dy = 0

Answer:

Disclaimer: There seems to be error in the given question.
 

Page No 21.139:

Question 49:

x2 dy + (x2xy + y2) dx = 0

Answer:

We have,x2dy+x2-xy+y2dy=0x2dy=xy-x2-y2dydydx=xy-x2-y2x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x2v-x2-x2v2x2v+xdvdx=v-1-v2xdvdx=-1-v2dv1+v2=-1xdxIntegrating both sides, we getdv1+v2dv=-1xdxtan-1 v=-log x+log Ctan-1 yx=logCxetan-1yx=CxC=x etan-1yx

Page No 21.139:

Question 50:

y-xdydx=b1+x2dydx

Answer:

We have,y-xdydx=b1+x2dydxy-b=bx2+xdydx1y-bdy=1bx2+xdxIntegrating both sides, we get1y-bdy=1bx2+xdx1y-bdy=1b1x2+1bxdx1y-bdy=1b1x2+1bx+14b2-14b2dx1y-bdy=1b1x+12b2-12b2dxlog y-b=12×12bblog x+12b-12bx+12b+12b+log Clog y-b=log bxbx+1+log Cy-b=Cbxbx+1Cbx=y-bbx+1x=ky-bbx+1, where k=1bC

Page No 21.139:

Question 51:

dydx+2y=sin 3x

Answer:

We have,dydx+2y=sin 3x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2 and Q=sin 3x.   I.F.=eP dx          =e2 dx         = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xsin 3x e2xdydx+2e2xy=e2xsin 3xIntegrating both sides with respect to x, we gety e2x=e2xsin 3x  dx+Cy e2x=I+C           .....1Where, I=e2xsin 3x dx           .....2I=e2xsin 3x dx-de2xdxsin 3x dxdxI=-e2xcos 3x3+23e2xcos 3x dxI=-e2xcos 3x3+23e2xcos 3x dx-de2xdxcos 3x dxdxI=-e2xcos 3x3+23e2xsin 3x3-23e2xsin 3x dxI=-e2xcos 3x3+2 e2xsin 3x9-49e2xsin 3x dxI=-e2xcos 3x3+2 e2xsin 3x9-49I             Using 213I9=-e2xcos 3x3+2 e2xsin 3x9I=9132 e2xsin 3x9-e2xcos 3x3I=e2x132 sin 3x-3 cos 3x           .....3From 1 and 3, we gety e2x=e2x132 sin 3x-3 cos 3x+Cy=31323sin 3x-cos 3x+Ce-2xHence, y=31323sin 3x-cos 3x+Ce-2x  is the required solution.

Page No 21.139:

Question 52:

dydx+y=4x

Answer:

We have,dydx+y=4x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=1 and Q=4x I.F.=eP dx          =e dx         = exMultiplying both sides of (1) by I.F.=ex, we getex dydx+y=ex4x exdydx+exy=ex4xIntegrating both sides with respect to x, we gety ex=4x ex dx+Cy ex=4xI exII dx+Cy ex=4xex dx-4ddxxex dxdx+Cy ex=4xex-4ex dx+Cy ex=4xex-4ex+Cy ex=4x-1ex+Cy=4x-1+Ce-xHence, y=4x-1+Ce-x  is the required solution.

Page No 21.139:

Question 53:

dydx+5y=cos 4x

Answer:

We have,dydx+5y=cos 4x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=15 and Q=cos 4x. I.F.=eP dx          =e 5dx         = e5xMultiplying both sides of (1) by I.F.=e5x, we gete5x dydx+5y=e5xcos 4x e5xdydx+5e5xy=e5xcos 4xIntegrating both sides with respect to x, we gety e5x=e5xcos 4x dx+Cy e5x=I+C       .....2Where,I=e5xcos 4x dx       .....3I=e5xcos 4x dx-de5xdxcos 4x dxdxI=e5xsin 4x4-54e5xsin 4x dxI=e5xsin 4x4-54e5xsin 4x dx-de5xdxsin 4x dxdxI=e5xsin 4x4-54-e5xcos 4x4+54e5xcos 4x dxI=e5xsin 4x4+5e5xcos 4x16-2516e5xcos 4x dxI=e5xsin 4x4+5e5xcos 4x16-2516I        From 34116I=e5xsin 4x4+5e5xcos 4x164116I=e5x164sin 4x+5cos 4xI=e5x414sin 4x+5cos 4x       .....4From 2 and 4 we gety e5x=e5x414sin 4x+5cos 4x+Cy=441sin 4x+54cos 4x+Ce-5xHence, y=441sin 4x+54cos 4x+Ce-5x  is the required solution.

Page No 21.139:

Question 54:

xdydx+x cos2yx=y

Answer:

We have,xdydx+x cos2yx=ydydx+cos2yx=yxdydx=yx-cos2yxPutting y=vx, we getdydx=v+xdvdx v+xdvdx=v-cos2vxdvdx=-cos2 vsec2 v dv=-1xdxIntegrating both sides, we getsec2 v dv=-1xdxtan v =-log x+Ctan yx =-log x+C

Page No 21.139:

Question 55:

cos2 xdydx+y=tan x

Answer:

We have,cos2 xdydx+y=tan xdydx+sec2 xy=tan xsec2 xComparing with dydx+Px=Q, we getP=sec2 x Q=tan xsec2 xNow,I.F.=esec2 x dx=etan xSo, the solution is given byy×etan x=tan xsec2 x×etan x dx + Cyetan x=I + C             .....1Now,I=tan xsec2 x×etan x dxPutting t=tan x, we getdt=sec2 x dx I=tI×etII dt     =t×etdt-dtdt×etdtdt     =tet-etdt     =tet-etI=tan x etan x-etan x=etan xtan x-1Putting the value of I in 1, we getyetan x=etan xtan x-1+ C

Page No 21.139:

Question 56:

x cos x dydx + y (x sin x + cos x) = 1

Answer:

We have,x cos xdydx+y x sin x+cos x=1dydx+tan x+1xy=1x cos xComparing with dydx+Px=Q, we getP=tan x+1x Q=1x cos xNow,I.F.=etan x+1x dx=elog x sec x=x sec xSo, the solution is given byxy sec x=sec2 x dx + Cxy sec x=tan x+C

Page No 21.139:

Question 57:

1+y2+x-e-tan-1 ydydx=0

Answer:

We have,
1+y2+x-e-tan-1 ydydx=0
dxdy=e-tan-1 y-x1+y2dxdy+x1+y2=e-tan-1 y1+y2

Comparing with dxdy+Px=Q, we getP=11+y2 Q=e-tan-1 y1+y2Now,I.F.=e11+y2dy=etan-1 ySo, the solution is given byx×etan-1 y=e-tan-1 y1+y2×etan-1 y dy + Cx×etan-1 y=11+y2 dy + Cxetan-1 y=tan-1 y + C

Page No 21.139:

Question 58:

y2+x+1ydydx=0

Answer:

We have,
y2+x+1ydydx=0
dydx=-y3xy+1dxdy=-xy+1y3dxdy=-xy2-1y3dxdy+xy2=-1y3
Comparing with dxdy+Px=Q, we getP=1y2Q=-1y3Now, I.F.=e1y2dy=e-1ySo, the solution is given byx×e-1y=-e-1y1y3 dy + Cxe-1y=I + C      .....1Now,I=-e-1y1y3 dyPutting t=1y, we getdt=-1y2dy I=tI×e-tII dt     =t×e-tdt-dtdt×e-tdtdt     =-tet+e-tdt     =-te-t-e-t I=-1ye-1y-e-1y=-e-1y1+1yPutting the value of I in 1, we getxe-1y=-e-1y1+1y+ Cx=-1+1y+ Ce1y



Page No 21.140:

Question 59:

2 cos xdydx+4y sin x = sin 2x, given that y = 0 when x = π3.

Answer:

We have,
2cos xdydx+4y sin x=sin 2x
dydx+4ysin x2 cos x=2sin x cos x2 cos xdydx+2y tan x=sin x
Comparing with dydx+Py=Q, we getP=2tan xQ=sin xNow,I.F.=e2tan x dx=e2logsec x=sec2 xSo, the solution is given byy×I.F.=Q×I.F. dx + Cy sec2 x=sin x sec2 x dx + Cy sec2 x=tan x sec x dx + Cy sec2 x=sec x  + Cy=cos x+C cos2x       .....1Now, When x=π3, y=0  0=cos π3  + C cos2 π30=12+C14C=-2Putting the value of C in 1, we gety=cos x-2cos2 x

Page No 21.140:

Question 60:

(1 + y2) dx = (tan−1 yx) dy

Answer:

We have,1+y2dx=tan-1y-xdydxdy=tan-1 y-x1+y2dxdy+x1+y2=tan-1 y1+y2
Comparing with dxdy+Px=Q, we getP=11+y2 Q=tan-1 y1+y2Now, I.F.=e11+y2dy=etan-1 ySo, the solution is given byx×etan-1 y=tan-1 y1+y2×etan-1 y dy + Cxetan-1 y=I + C         .....1Now, I=tan-1 y1+y2×etan-1 y dyPutting t=tan-1 y, we getdt=11+y2dy I=tI×etII dt     =t×etdt-dtdt×etdtdt     =tet-etdt     =tet-et I=tan-1y etan-1 y-etan-1 y    =etan-1 ytan-1 y-1Putting the value of I in 1, we getxetan-1 y=etan-1 ytan-1 y-1+ C

Page No 21.140:

Question 61:

dydx + y tan x = xn cos x, n ≠ − 1

Answer:

We have,dydx+y tan x=xn cos x
Comparing with dydx+Py=Q, we getP=tan x Q=xn cos xNow, I.F.=etan x dx=elogsec x=sec xSo, the solution is given byy×I.F.=Q×I.F. dx + Cy sec x=xn cos x sec x dx + Cy sec x=xn dx + Cy sec x=xn+1n+1  + C

Page No 21.140:

Question 62:

Find the general solution of the differential equation dydx=x+12-y, y2.

Answer:

We have,dydx=x+12-y2-ydy=x+1dxIntegrating both sides, we get2-ydy=x+1dx2y-y22=x22+x+C1x22+x+C1-2y+y22=0x2+2x+y2+2C1-4y=0x2+y2+2x-4y+C=0         Where, C=2C1

Page No 21.140:

Question 63:

Find the particular solution of the differential equation dydx=-4xy2 given that y = 1, when x = 0.

Answer:

We have,dydx=-4xy21y2dy=-4x dxIntegrating both sides, we get1y2dy=-4x dx-1y=-2x2+C      .....1Now, When x=0, y=1  -1=0+CC=-1Putting the value of C in 1, we get-1y=-2x2-11y=2x2+1y=12x2+1

Page No 21.140:

Question 64:

For each of the following differential equations, find the general solution:
(i) dydx=1-cos x1+cos x

(ii) dydx=4-y2, -2<y<2

(iii) dydx=1+x21+y2

(iv) y log y dxx dy = 0

(v) dydx=sin-1x

(vi) dydx+y=1

Answer:

i We have,dydx=1-cos x1+cos xdydx=2sin2 x22cos2 x2dydx=tan2 x2dy=tan2 x2dxIntegrating both sides, we getdy=tan2 x2dxdy=sec2 x2-1dxy= 2 tan x2-x+C

ii We have,dydx=4-y214-y2dy=dxIntegrating both sides, we get14-y2dy=dxsin-1 y2=x+Cy2=sin x+Cy=2sin x+C

iii We have,dydx=1+x21+y211+y2dy=1+x2dxIntegrating both sides, we get11+y2dy=1+x2dxtan-1 y=x+x33+C

iv We have,y log y dx-x dy=0y log y dx=x dy1xdx=1y log ydy1y log ydy=1xdxIntegrating both sides, we get1y log ydy=1xdx       .....1Putting log y=t1ydy=dtTherefore 1 becomes1tdt=1xdxlog t=log x + log Clog log y=log x + log Clog log y=log Cxlog y=Cxy=eCx

v We have,dydx=sin-1xdy=sin-1xdxIntegrating both sides, we getdy=sin-1xdxdy=1II×sin-1xI dx dy=sin-1x1 dx-ddxsin-1x1 dxdxy=x sin-1x-x1-x2dxPutting t2=1-x2, we get2t dt=-2x dx-t dt=x dx y=x sin-1x+dty=x sin-1x+t+Cy=x sin-1x+1-x2+C

vi We have,dydx+y=1dydx=1-y11-ydy=dxIntegrating both sides, we get-1y-1dy=dx1y-1dy=-dxlog y-1=-x+log Clog y-1-log C=-xlog y-1C=-xy-1C=e-xy=1+Ce-x

Page No 21.140:

Question 65:

For each of the following differential equations, find a particular solution satisfying the given condition:
(i) xx2-1dydx=1, y=0 when x=2

(ii) cosdydx=a, y=1 when x=0

(iii) dydx=y tan x, y=1 when x=0

Answer:

i We have,xx2-1dydx=1 dydx=1xx2-1dy=1xx2-1dxIntegrating both sides, we getdy=1xx2-1dxy=1xx2-1dx+Cy=1xx+1x-1dx+C     .....1Let 1xx+1x-1=Ax+Bx+1+Cx-11=Ax+1x-1+Bxx-1+Cxx+11=Ax2-1+Bx2-x+Cx2+x1=x2A+B+C+x-B+C-AComparing both sides, we get-A=1          .....2-B+C=0     .....3A+B+C=0     .....4Solving 2, 3 and 4, we getA=-1B=12C=121xx+1x-1=-1x+12x+1+12x-1Now, 1 becomesy=-1x+12x+1+12x-1dx+Cy=-1xdx+121x-1dx+121x-1dxy=-log x+12log x-1+12log x+1+Cy=12log x-1+12log x+1-log x+CGiven:y2=0 0=12log 2-1+12log 2+1-log 2+CC=log 2-12log 3Substituting the value of C, we gety=12log x-1+12log x+1-log x+log 2-12log 32y=log x-1+log x+1-2log x+2log 2-log 32y=log x-1+log x+1-log x2+log 4-log 32y=logx-1x+1x2-log3-log4y=12logx2-1x2-12log 34

ii We have,cos dydx=a  dydx=cos-1 ady=cos-1 a dxIntegrating both sides, we getdy=cos-1 a dxy=x cos-1 a+CNow, When x=0, y=1  1=0+CC=1Putting the value of C in 1, we gety=x cos-1 a+1cosy-1x=a

iii We have,dydx=y tan x1ydy=tan x dxIntegrating both sides, we get1ydy=tan x dxlog y=log sec x+C     ....1Now, When x=0, y=1  log 1=log 1+CC=0Putting the value of C in 1, we getlog y=log sec xy=sec x

Page No 21.140:

Question 66:

Solve the each of the following differential equations:
(i) x-ydydx=x+2y

(ii) x cosyxdydx=y cosyx+x

(iii) y dx + x log yx dy − 2x dy = 0

(iv) dydx-y=cos x

(v) xdydx+2y=x2, x0

(vi) dydx+2y=sin x

(vii) dydx+3y=e-2x

(viii) dydx+yx=x2

(ix) dydx+sec x y=tan x

(x) xdydx+2y=x2 log x

(xi) x log xdydx+y=2xlog x

(xii) (1 + x2) dy + 2xy dx = cot x dx

(xiii) x+ydydx=1

(xiv) y dx + (xy2) dy = 0

(xv) x+3y2dydx=y

Answer:

i We have,x-ydydx=x+2ydydx=x+2yx-y     .....1Clearly this is a homogeneous equation,Putting y=vxdydx=v+xdvdxSubstituting y=vx and dydx=v+xdvdx 1 becomes,v+xdvdx=x+2vxx-vxv+xdvdx=1+2v1-vxdvdx=1+2v1-v-vxdvdx=1+2v-v+v21-vxdvdx=v2+v+11-v1-vv2+v+1dv=1xdx-vv2+v+1+1v2+v+1dv=1xdx-12×2v+1-1v2+v+1+1v2+v+1dv=1xdx-12×2v+1v2+v+1+12×1v2+v+1+1v2+v+1dv=1xdx-12×2v+1v2+v+1+32×1v2+v+1dv=1xdx-12×2v+1v2+v+1+32×1v2+v+14+34dv=1xdx-12×2v+1v2+v+1+32×1v+122+322dv=1xdxIntegrating both sides, we get-12×2v+1v2+v+1+32×1v+122+322dv=1xdx-122v+1v2+v+1dv+321v+122+322dv=1xdx-12log v2+v+1+32×132tan-1v+1232=log x+C-12log yx2+yx+1+32×132tan-1yx+1232=log x+C-12log y2+xy+x2x2+3tan-12y+x3x=log x+C-12log y2+xy+x2+12log x2+3tan-12y+x3x=log x+C-12log y2+xy+x2+log x+3tan-12y+x3x=log x+C-12log y2+xy+x2+3tan-12y+x3x=Clog y2+xy+x2-23tan-12y+x3x=-2Clog y2+xy+x2=23tan-12y+x3x-2Clog y2+xy+x2=23tan-12y+x3x+k         Where, k=-2C

ii We have,x cos yxdydx=y cosyx+xdydx=y cosyx+xx cos yx     .....1Clearly this is a homogeneous equation,Putting y=vxdydx=v+xdvdxSubstituting y=vx and dydx=v+xdvdx in 1 we getdydx=y cosyx+xx cos yxv+xdvdx=vx cos v+xx cos vv+xdvdx=v cos v+1cos vxdvdx=v cos v+1cos v-vxdvdx=v cos v+1- v cos vcos vxdvdx=1cos vcos v dv=1xdxIntegrating both sides, we getcos v dv=1xdxsin v=log x+log Csin yx=log Cx

iii We have,y dx+x log yxdy-2x dy=0x log yxdy-2x dy=-y dxlog yx-2x dy=-y dxdydx=-ylog yx-2xdydx=yx2-log yx     .....1Clearly this is a homogenous equation,Putting y=vxdydx=v+xdvdxSubstituting y=vx and dydx=v+xdvdx in 1 we getv+xdvdx=v2-log vxdvdx=v2-log v-vxdvdx=v-2v+v log v2-log vxdvdx=-v+v log v2-log v2-log v-v+v log vdv=1xdxlog v-2v log v-vdv=-1xdxlog v-1-1v log v-1dv=-1xdxlog v-1v log v-1dv-1v log v-1dv=-1xdx1vdv-1v log v-1dv=-1xdxIntegrating both sides we get1vdv-1v log v-1dv=-1xdxlog v-I =-log x-log C     .....2Where,I=1v log v-1dvPuting log v=t1vdv=dtI=1t-1dtI=log t-1I=log log v-1     .....3From 2 and 3 we getlog v-log log v-1 =-log x-log Clog vlog v-1 =-log Cxvlog v-1=1Cxlog v-1=vCxlog yx-1=Cy


iv We have,dydx-y=cos xComparing with dydx+Py=Q, we getP=-1 Q=cos xNow, I.F.=e-1dx =e-xSolution is given by,y×I.F.=cos x×I.F. dx+Cye-x=e-x cos x dx+Cye-x=I+C     .....1Where,I=e-xIIcos x Idx     .....2 I=cos xe-x dx-ddxcos xe-x dxdx I=-cos x e-x-sin x e-x dx I=-cos x e-x-sin xI e-xII dx I=-cos x e-x-sin xe-x dx+ddxsin xe-x dxdx I=-c