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Mathematics
Problem
The solution of the differential equation
x
d
y
d
x
+
2
y
=
x
2
x \frac{dy}{dx} + 2y = x^2
x
d
x
d
y
+
2
y
=
x
2
(
x
≠
0
x \neq 0
x
=
0
) with
y
(
1
)
=
1
y(1) = 1
y
(
1
)
=
1
, is [JEE-MAIN ONLINE-2019] (1)
y
=
x
3
5
+
1
5
x
2
y = \frac{x^3}{5} + \frac{1}{5x^2}
y
=
5
x
3
+
5
x
2
1
(2)
y
=
4
5
x
3
+
1
5
x
2
y = \frac{4}{5}x^3 + \frac{1}{5x^2}
y
=
5
4
x
3
+
5
x
2
1
(3)
y
=
3
4
x
2
+
1
4
x
2
y = \frac{3}{4}x^2 + \frac{1}{4x^2}
y
=
4
3
x
2
+
4
x
2
1
(4)
y
=
x
2
4
+
3
4
x
2
y = \frac{x^2}{4} + \frac{3}{4x^2}
y
=
4
x
2
+
4
x
2
3
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