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Exams
Joint Entrance Examination
Mathematics
Differential Equations
Question 321
Question 321 of 490
Easy
MCQ
The solution of
d
y
d
x
=
x
+
y
x
\dfrac{dy}{dx}=\dfrac{x+y}{x}
d
x
d
y
=
x
x
+
y
,
y
(
1
)
=
1
y(1)=1
y
(
1
)
=
1
, is:
A
y
=
ln
x
+
x
y=\ln x+x
y
=
ln
x
+
x
B
y
=
x
ln
x
+
x
2
y=x\ln x+x^2
y
=
x
ln
x
+
x
2
C
y
=
x
e
(
x
−
1
)
y=xe^{(x-1)}
y
=
x
e
(
x
−
1
)
D
y
=
x
ln
x
+
x
y=x\ln x+x
y
=
x
ln
x
+
x
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Solutions · 1
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Official solution
With
y
=
v
x
y=vx
y
=
v
x
:
v
+
x
d
v
d
x
=
1
+
v
⟹
d
v
=
d
x
x
⟹
v
=
ln
x
+
c
v+x\dfrac{dv}{dx}=1+v \implies dv=\dfrac{dx}{x} \implies v=\ln x+c
v
+
x
d
x
d
v
=
1
+
v
⟹
d
v
=
x
d
x
⟹
v
=
ln
x
+
c
. Using
y
(
1
)
=
1
y(1)=1
y
(
1
)
=
1
:
c
=
1
c=1
c
=
1
, so
y
=
x
ln
x
+
x
y=x\ln x+x
y
=
x
ln
x
+
x
.
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