JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Differential Equations

The solution of the differential equation

x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)

with

y (1) = 1

, is:

Answer: (B)

Solution

Write the equation as

\frac{d y}{d x} + \frac{2}{x} y = x

, a linear equation with integrating factor

e^{\int 2/ x d x} = x^{2}

. Multiplying through, the left side becomes

\frac{d}{d x} (x^{2} y) = x^{3}

, so

x^{2} y = \frac{x ^{4}}{4} + C

. The condition

y (1) = 1

gives

1 = \frac{1}{4} + C

, so

C = \frac{3}{4}

and

y = \frac{x ^{2}}{4} + \frac{3}{4 x ^{2}}

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