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EAMCET Engineering 2002 Solved Paper
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© examsnet.com
Question : 158
Total: 200
The solution of
d
y
d
x
+
1
3
y
=
1
is
y
=
3
+
c
e
x
∕
3
y
=
3
+
c
e
−
x
∕
3
3
y
=
c
+
e
x
∕
3
y
2
+
x
+
x
2
+
2
=
c
e
2
x
Validate
Solution:
We have,
d
y
d
x
+
y
3
=
1
⇒
d
y
=
−
(
y
−
3
)
3
dx
⇒
d
y
y
−
3
=
−
1
3
dx
⇒
log
(
y
−
3
)
=
−
1
3
x
+
log
c
⇒
y
−
3
=
c
e
−
x
∕
3
⇒
y
=
3
+
c
e
−
x
∕
3
© examsnet.com
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