JEE Main 2026 (Online) 2nd April Evening ShiftQuestion 1
Let $x=x(y)$ be the solution of the differential equation $2y^{2}\frac{dx}{dy}-2xy+x^{2}=0$, $y>1$, $x(e)=e$. Then $x(e^{2})$ is equal to :
Differential Equation
Solution
Concept. The equation is homogeneous in $x$ and $y$; put $x=vy$. Formulas used. (1) $x=vy\Rightarrow\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$Step 1 - Substitute. By (1), $$2y^2\left(v+yv'\right)-2vy^2+v^2y^2=0\ \Longrightarrow\ 2y\,v'=-v^2.$$ Step 2 - Separate and integrate. $$\frac{dv}{v^2}=-\frac{dy}{2y}\ \Longrightarrow\ \frac1v=\frac12\ln y+C.$$ Step 3 - Constant. At $y=e$, $v=\frac xy=1$: $1=\frac12+C$, so $C=\frac12$ and $\frac1v=\frac{\ln y+1}{2}$. Step 4 - At $y=e^2$. $\frac1v=\frac32$, $v=\frac23$, so $x=\frac23e^2$. Answer: (B)