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

y = y (x)

is solution of the differential equation

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

satisfying

y (1) = 1

, then

y (\frac{1}{2})

is equal to

Answer: (C)

Solution

Put the equation in standard linear form

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

, whose integrating factor is

x^{2}

. Then

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

, so

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

. The condition

y (1) = 1

gives

C = \frac{3}{4}

, hence

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

. At

x = \frac{1}{2}

y = \frac{\frac{1}{16} + 3}{4 \cdot \frac{1}{4}} = \frac{49}{16}

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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