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Joint Entrance Examination
Mathematics
Differential Equations
Question 262
Question 262 of 490
Medium
MCQ
The solution of
d
y
d
x
−
y
+
3
x
log
e
(
y
+
3
x
)
+
3
=
0
\dfrac{dy}{dx}-\dfrac{y+3x}{\log_e(y+3x)}+3=0
d
x
d
y
−
lo
g
e
(
y
+
3
x
)
y
+
3
x
+
3
=
0
is (C constant):
A
x
−
1
2
(
log
e
(
y
+
3
x
)
)
2
=
C
x-\dfrac12(\log_e(y+3x))^2=C
x
−
2
1
(
lo
g
e
(
y
+
3
x
)
)
2
=
C
B
x
−
log
e
(
y
+
3
x
)
=
C
x-\log_e(y+3x)=C
x
−
lo
g
e
(
y
+
3
x
)
=
C
C
y
+
3
x
−
1
2
(
log
e
x
)
2
=
C
y+3x-\dfrac12(\log_ex)^2=C
y
+
3
x
−
2
1
(
lo
g
e
x
)
2
=
C
D
x
−
2
log
e
(
y
+
3
x
)
=
C
x-2\log_e(y+3x)=C
x
−
2
lo
g
e
(
y
+
3
x
)
=
C
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Solutions · 1
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Official solution
Let
t
=
y
+
3
x
t=y+3x
t
=
y
+
3
x
:
d
t
d
x
=
t
ln
t
⟹
ln
t
t
d
t
=
d
x
⟹
(
ln
t
)
2
2
=
x
−
C
\dfrac{dt}{dx}=\dfrac{t}{\ln t} \implies \dfrac{\ln t}{t}\,dt=dx \implies \dfrac{(\ln t)^2}{2}=x-C
d
x
d
t
=
ln
t
t
⟹
t
ln
t
d
t
=
d
x
⟹
2
(
ln
t
)
2
=
x
−
C
. \[x-\frac12(\ln(y+3x))^2 = C.\]
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