JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

(x^{2} - 3 y^{2}) d x + 3 x y d y = 0, y (1) = 1

. Then

6 y^{2} (e)

is equal to

Answer: (C)

Solution

Dividing through,

\frac{d y}{d x} = \frac{3 y ^{2} - x ^{2}}{3 x y} = \frac{y}{x} - \frac{x}{3 y}

, a homogeneous equation. Setting

y = v x

gives

v + x \frac{d v}{d x} = v - \frac{1}{3 v}

, so

v d v = - \frac{d x}{3 x}

. Integrating,

\frac{v ^{2}}{2} = - \frac{1}{3} ln x + C

, and

y (1) = 1

(so

v = 1

x = 1

) gives

C = \frac{1}{2}

. Hence

y^{2} = x^{2} (1 - \frac{2}{3} ln x)

. At

x = e

y^{2} (e) = e^{2} (1 - \frac{2}{3}) = \frac{e ^{2}}{3}

, so

6 y^{2} (e) = 2 e^{2}

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