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UPSC NDA Math Model Paper 3
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© examsnet.com
Question : 43
Total: 120
A general solution of
log
(
d
y
d
x
)
=
2
x
+
3
y
,
y
(
0
)
=
0
is
2
e
−
3
y
+
3
e
2
x
=
−
5
2
e
−
3
y
+
3
e
2
x
=
5
3
e
−
3
y
+
2
e
2
x
=
5
None of the above
Validate
Solution:
Given differential equation is
log
(
d
y
d
x
)
=
2
x
+
3
y
⇒
d
y
d
x
=
e
(
2
x
+
3
y
)
⇒
d
y
d
x
=
e
2
x
e
3
y
⇒
d
y
e
3
y
=
e
2
x
d
x
Integrating both sides, we get
⇒
∫
e
−
3
y
d
y
=
∫
e
2
x
d
x
⇒
e
−
3
y
−
3
=
e
2
x
2
+
c
⇒
2
e
−
3
y
=
−
3
e
2
x
−
6
c
.......(1)
By using y = 0 when x = 0, we get
⇒ 2 = -3 – 6c
∴ c = -5/6
Put the value of
c
in equation
1
st
⇒
2
e
−
3
y
=
−
3
e
2
x
−
6
(
−
5
∕
6
)
⇒
2
e
−
3
y
=
−
3
e
2
x
+
5
⇒
2
e
−
3
y
+
3
e
2
x
=
5
© examsnet.com
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