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Shanaya
Subject: Maths
, asked on 19/3/18
In example 7 how did we arrive at the last step? or given
Answer
2
Shanaya
Subject: Maths
, asked on 18/3/18
In the second step how did we get -4 as power In RHS?
Answer
1
Archit
Subject: Maths
, asked on 18/3/18
Find the differential eq. .. As the general solution
Q.11. Which of the following differential equations has
y
=
c
1
e
x
+
c
2
e
-
x
as the general solution?
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 18/3/18
Q If
y
=
e
a
cos
-
1
x
,
t
h
e
n
s
h
o
w
t
h
a
t
(
1
-
x
2
)
y
2
-
x
y
1
-
a
2
y
=
0
Answer
1
Indu Nair
Subject: Maths
, asked on 18/3/18
Please solve
Q.
d
v
d
t
=
g
cos
α
+
k
v
;
k
0
=
0
Answer
0
Shubhrajyoti Ghosh
Subject: Maths
, asked on 17/3/18
No links please
Q. iii) Solve :
x
y
2
-
e
1
x
3
d
x
-
x
2
y
d
y
=
0
; given y = 0 when x = 1.
Ans:
y
2
=
2
2
e
x
1
3
-
e
x
2
Answer
1
Manasi Mujumdar
Subject: Maths
, asked on 16/3/18
Solve the following
Differential equation
-
y
+
3
x
2
d
x
d
y
=
x
Answer
1
Bhanvi
Subject: Maths
, asked on 16/3/18
In a college hostel accommodating 1000 students, one of the hostellers came in carrying a flu virus, and the hostel was isolated. If the rate at which the virus spreads is assumed to be proportional to the product of the number N of infected students and the number of non-infected students, and if infected students are 50 after 4 days then show that more than 95% of the hostellers will be infected after 10 days. If Shyam was the first student to be infected what precautions he should have initiated to avoid this situation.
Answer
1
Aishwarya Trivedi
Subject: Maths
, asked on 15/3/18
in this soln part i didnt get after (((log(1-y)=x+logc)))) . in the next line why there is a minus sign before log(1-y)??? because when logc is bring to the left side it should be -ve but why log(1-y)is -ve?:
Now, Integrating both sides, we Get:
∫
d
y
1
-
y
=
∫
d
x
⇒
log
1
-
y
=
x
+
log
C
⇒
-
log
C
-
log
1
-
y
=
x
⇒
log
C
1
-
y
=
-
x
⇒
C
(
1
-
y
)
=
e
-
x
⇒
1
-
y
=
1
c
e
-
x
⇒
y
=
1
-
1
c
e
-
x
⇒
y
=
A
e
-
x
W
h
e
r
e
A
=
-
1
c
∫
d
y
1
-
y
=
∫
d
x
⇒
log
1
-
y
=
x
+
log
C
⇒
-
log
C
-
log
1
-
y
=
x
⇒
log
C
1
-
y
=
-
x
⇒
C
(
1
-
y
)
=
e
-
x
⇒
1
-
y
=
1
c
e
-
x
⇒
y
=
1
-
1
c
e
-
x
⇒
y
=
A
e
-
x
W
h
e
r
e
A
=
-
1
c
Answer
1
Jeneeta Eliza John
Subject: Maths
, asked on 15/3/18
a
+
x
d
y
d
x
+
x
y
=
0
Answer
2
Shubhrajyoti Ghosh
Subject: Maths
, asked on 15/3/18
No links please
Q. Solve :
y
2
+
x
-
1
y
d
y
d
x
=
0
Ans :
x
-
1
y
-
1
=
c
e
1
y
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 15/3/18
No links please
Q.(iii) Solve :
d
y
d
x
-
y
x
+
C
o
s
c
e
y
x
=
0
; given y = 0 when x = 1.
Ans :
log
x
+
1
=
cos
y
x
Answer
1
Shubhrajyoti Ghosh
Subject: Maths
, asked on 15/3/18
No links please:
Q. Prove that
d
2
y
d
x
2
=
x
1
y
2
-
y
1
x
2
x
1
3
Where
x
1
=
d
f
d
t
,
y
1
=
d
g
d
t
,
x
2
=
d
2
f
d
t
2
,
y
2
=
d
2
g
d
t
2
Answer
0
Megha Prasadan
Subject: Maths
, asked on 15/3/18
Q. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Answer
1
Devi Das
Subject: Maths
, asked on 14/3/18
Q). Solve:
d
y
d
x
=
2
x
-
3
y
+
4
3
x
+
4
y
-
5
Answer
1
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What are you looking for?
Q.11. Which of the following differential equations has as the general solution?
Q.
Q. iii) Solve : ; given y = 0 when x = 1.
Ans:
Now, Integrating both sides, we Get:
Q. Solve :
Ans :
Q.(iii) Solve : ; given y = 0 when x = 1.
Ans :
Q. Prove that