JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Differential Equations

Let

x = x (y)

be the solution of the differential equation

2 y^{2} \frac{d x}{d y} - 2 x y + x^{2} = 0, y > 1, x (e) = e

. Then

x (e^{2})

is equal to:

Answer: (B)

Solution

Write the equation as

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

and put

x = v y

, so

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

. This gives

y \frac{d v}{d y} = - \frac{v ^{2}}{2}

. Separating,

\frac{d v}{v ^{2}} = - \frac{d y}{2 y}

, so

\frac{1}{v} = \frac{1}{2} ln y + C

, i.e.

\frac{y}{x} = \frac{1}{2} ln y + C

. The condition

x (e) = e

gives

1 = \frac{1}{2} + C

, so

C = \frac{1}{2}

. At

y = e^{2}

\frac{e ^{2}}{x} = \frac{1}{2} (2) + \frac{1}{2} = \frac{3}{2}

, hence

x (e^{2}) = \frac{2}{3} e^{2}

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