JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Differential Equations

Let

y (x)

be the solution of the differential equation

x^{2} \frac{d y}{d x} + x y = x^{2} + y^{2}, x > \frac{1}{e},

satisfying

y (1) = 0

. Then the value of

2 \frac{( y ( e ) ) ^{2}}{y ( e ^{2} )}

is ____.

Answer: 0.75

Solution

Substitute

y = v x

, so

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

.

The equation becomes:

x^{2} (v + x \frac{d v}{d x}) + x^{2} v = x^{2} (1 + v^{2})

x \frac{d v}{d x} = 1 + v^{2} - 2 v = (v - 1)^{2}

Separating variables:

\int \frac{d v}{( v - 1 ) ^{2}} = \int \frac{d x}{x}

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

Substituting back

v = \frac{y}{x}

\frac{x}{x - y} = ln x + C

Using

y (1) = 0

\frac{1}{1} = 0 + C

, so

C = 1

.

Thus

\frac{x}{x - y} = ln (e x)

.

For

x = e

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

, so

e = 2 (e - y (e))

, giving

y (e) = \frac{e}{2}

.

For

x = e^{2}

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

, so

e^{2} = 3 (e^{2} - y (e^{2}))

, giving

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

.

Therefore:

2 \frac{( y ( e ) ) ^{2}}{y ( e ^{2} )} = 2 \cdot \frac{( e /2 ) ^{2}}{2 e ^{2} /3} = 2 \cdot \frac{e ^{2} /4}{2 e ^{2} /3} = 2 \cdot \frac{3}{8} = \frac{3}{4} = 0.75

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer