Let the solution curve of the differential equation $x dy-y dx=\sqrt{x^{2}+y^{2}} dx, x>0, y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Differential Equations

Let the solution curve of the differential equation

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

, be

y = y (x)

. Then

y (3)

is equal to

Answer: (A)

Putting

y = v x

converts

x d y - y d x = x^{2} + y^{2} d x

into

x^{2} d v = x 1 + v^{2} d x

(using

x > 0

), i.e.

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

. Integrating gives

sinh^{- 1} v = ln x + C

, and

y (1) = 0

forces

C = 0

. Hence

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

, so

y = \frac{x ^{2} - 1}{2}

. At

x = 3

y (3) = \frac{9 - 1}{2} = 4

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

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