JEE Main 2019 — Math solution | Fundamenthol

JEE Main2019Math

Q.

The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)$ with y(1) = 1, is :

A
$y = \frac{4}{5} x^{3} + \frac{1}{5 x ^{2}}$
B
$y = \frac{x ^{3}}{5} + \frac{1}{5 x ^{2}}$
C
$y = \frac{x ^{2}}{4} + \frac{3}{4 x ^{2}}$
D
$y = \frac{3}{4} x^{2} + \frac{1}{4 x ^{2}}$

Solution

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