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Joint Entrance Examination
Mathematics
Differential Equations
Question 270
Question 270 of 490
Easy
MCQ
If
y
=
y
(
x
)
y=y(x)
y
=
y
(
x
)
solves
x
d
y
d
x
+
2
y
=
x
2
x\dfrac{dy}{dx}+2y=x^2
x
d
x
d
y
+
2
y
=
x
2
,
y
(
1
)
=
1
y(1)=1
y
(
1
)
=
1
, then
y
(
1
2
)
y\left(\dfrac12\right)
y
(
2
1
)
is:
A
7
64
\dfrac{7}{64}
64
7
B
1
4
\dfrac14
4
1
C
49
16
\dfrac{49}{16}
16
49
D
13
16
\dfrac{13}{16}
16
13
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Official solution
I.F.
=
x
2
=x^2
=
x
2
.
y
x
2
=
x
4
4
+
C
yx^2 = \dfrac{x^4}{4}+C
y
x
2
=
4
x
4
+
C
. Using
y
(
1
)
=
1
y(1)=1
y
(
1
)
=
1
:
C
=
3
4
C=\dfrac34
C
=
4
3
. At
x
=
1
/
2
x=1/2
x
=
1/2
:
y
⋅
1
4
=
1
64
+
3
4
⟹
y
=
49
16
y\cdot\dfrac14 = \dfrac{1}{64}+\dfrac34 \implies y=\dfrac{49}{16}
y
⋅
4
1
=
64
1
+
4
3
⟹
y
=
16
49
.
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