Q30 / 75 · JEE Main 2026 (22 January Shift 1) · Mathematics
Let the solution curve of the differential equation $x d y-y d x=\sqrt{x^{2}+y^{2}} d x, x>0$, $y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to
- A1
- Correct answerB4
- C2
- D6
Solution
$xdy - ydx = \sqrt{x^2+y^2}\,dx$.
Put $y = vx$: $x^2 dv = x\sqrt{1+v^2}\,dx$.
$\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$.
Integrating: $\ln(v+\sqrt{1+v^2}) = \ln x + C$.
$y(1) = 0 \Rightarrow v = 0$ at $x = 1$: $C = 0$.
$v + \sqrt{1+v^2} = x \Rightarrow y + \sqrt{x^2+y^2} = x^2$.
At $x = 3$: $y + \sqrt{9+y^2} = 9$
$\Rightarrow 9+y^2 = (9-y)^2$
$\Rightarrow 18y = 72$
$\Rightarrow y = 4$.
Put $y = vx$: $x^2 dv = x\sqrt{1+v^2}\,dx$.
$\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$.
Integrating: $\ln(v+\sqrt{1+v^2}) = \ln x + C$.
$y(1) = 0 \Rightarrow v = 0$ at $x = 1$: $C = 0$.
$v + \sqrt{1+v^2} = x \Rightarrow y + \sqrt{x^2+y^2} = x^2$.
At $x = 3$: $y + \sqrt{9+y^2} = 9$
$\Rightarrow 9+y^2 = (9-y)^2$
$\Rightarrow 18y = 72$
$\Rightarrow y = 4$.
Hence, the correct answer is option: 'B'.