JEE Main 2026 — 22 January, Morning Shift — Question 18

Given: $x\,dy - y\,dx = \sqrt{x^2+y^2}\,dx,\; x>0,\; y(1)=0$. Divide by $x^2$: $\frac{x\,dy - y\,dx}{x^2} = \frac{\sqrt{x^2+y^2}}{x^2}\,dx$. Left side is $d\left(\frac{y}{x}\right)$, right side: $\sqrt{1+\left(\frac{y}{x}\right)^2}\cdot\frac{1}{x}\,dx$. Thus $d\left(\frac{y}{x}\right) = \sqrt{1+\left(\frac{y}{x}\right)^2}\cdot\frac{dx}{x}$. Integrate: $\int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1+\left(\frac{y}{x}\right)^2}} = \int \frac{dx}{x}$. $\ln\left(\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}\right) = \ln x + \ln k = \ln(kx)$. Hence $\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}} = kx \Rightarrow y+\sqrt{x^2+y^2} = kx^2$. Using $y(1)=0$: $0+\sqrt{1+0}=k\cdot1 \Rightarrow k=1$. So $y+\sqrt{x^2+y^2}=x^2$. For $x=3$: $y+\sqrt{9+y^2}=9$.

Solve: $\sqrt{9+y^2}=9-y$. Squaring: $9+y^2 = 81-18y+y^2 \Rightarrow 18y=72 \Rightarrow y=4$.