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Exams
Joint Entrance Examination
Mathematics
Differential Equations
Question 473
Question 473 of 490
Medium
MCQ
Let
y
=
y
(
x
)
y=y(x)
y
=
y
(
x
)
be the solution of the differential equation \[\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0\] and \[y\left(e^{-1}\right)=0\]. Then,
y
(
e
)
y(e)
y
(
e
)
is equal to
A
\[-\frac{3}{\mathrm{e}}\]
B
\[-\frac{3}{2 \mathrm{e}}\]
C
\[-\frac{2}{3 \mathrm{e}}\]
D
\[-\frac{2}{\mathrm{e}}\]
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Solutions · 1
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Official solution
\[
(
2
x
log
x
)
d
y
d
x
+
2
y
=
3
x
log
x
d
y
d
x
+
y
x
log
x
=
3
2
x
2
I
F
=
e
∫
1
x
log
x
=
e
log
(
log
x
)
=
log
x
y
×
log
x
=
∫
log
x
×
3
2
x
2
d
x
+
C
y
log
x
=
3
2
[
−
log
x
x
−
1
x
]
+
C
y
(
e
−
1
)
=
0
⇒
0
=
3
2
[
e
−
e
]
+
C
C
=
0
y
log
x
=
−
3
2
[
log
x
+
1
x
]
y
(
e
)
→
y
(
e
)
×
1
=
−
3
2
[
1
+
1
e
]
=
−
3
e
\begin{aligned} & (2 x \log x) \frac{d y}{d x}+2 y=\frac{3}{x} \log x \\ & \frac{d y}{d x}+\frac{y}{x \log x}=\frac{3}{2 x^2} \\ & I F=e^{\int \frac{1}{x \log x}}=e^{\log (\log x)}=\log x \\ & y \times \log x=\int \log x \times \frac{3}{2 x^2} d x+C \\ & y \log x=\frac{3}{2}\left[\frac{-\log x}{x}-\frac{1}{x}\right]+C \\ & y\left(e^{-1}\right)=0 \\ & \Rightarrow 0=\frac{3}{2}[e-e]+C \\ & C=0 \\ & y \log x=\frac{-3}{2}\left[\frac{\log x+1}{x}\right] \\ & y(e) \rightarrow y(e) \times 1=\frac{-3}{2}\left[\frac{1+1}{e}\right]=-\frac{3}{e} \end{aligned}
(
2
x
lo
g
x
)
d
x
d
y
+
2
y
=
x
3
lo
g
x
d
x
d
y
+
x
lo
g
x
y
=
2
x
2
3
I
F
=
e
∫
x
l
o
g
x
1
=
e
l
o
g
(
l
o
g
x
)
=
lo
g
x
y
×
lo
g
x
=
∫
lo
g
x
×
2
x
2
3
d
x
+
C
y
lo
g
x
=
2
3
[
x
−
lo
g
x
−
x
1
]
+
C
y
(
e
−
1
)
=
0
⇒
0
=
2
3
[
e
−
e
]
+
C
C
=
0
y
lo
g
x
=
2
−
3
[
x
lo
g
x
+
1
]
y
(
e
)
→
y
(
e
)
×
1
=
2
−
3
[
e
1
+
1
]
=
−
e
3
\]
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