JEE Main 2026 — Mathematics Calculus

The given differential equation is $2y^2 \dfrac{dx}{dy} - 2xy + x^2 = 0$2y2dydx​−2xy+x2=0.

Dividing by $2y^2$2y2, we get:

$$\dfrac{dx}{dy} - \dfrac{x}{y} + \dfrac{x^2}{2y^2} = 0$$dydx​−yx​+2y2x2​=0

This is a homogeneous differential equation. Let $x = vy$x=vy, then $\dfrac{dx}{dy} = v + y \dfrac{dv}{dy}$dydx​=v+ydydv​.

Substituting these into the equation:

$$v + y \dfrac{dv}{dy} - v + \dfrac{v^2}{2} = 0$$v+ydydv​−v+2v2​=0

$$y \dfrac{dv}{dy} = -\dfrac{v^2}{2}$$ydydv​=−2v2​

Separating the variables:

$$\dfrac{dv}{v^2} = -\dfrac{dy}{2y}$$v2dv​=−2ydy​

Integrating both sides:

$$\int v^{-2} dv = -\dfrac{1}{2} \int \dfrac{dy}{y}$$∫v−2dv=−21​∫ydy​

$$-\dfrac{1}{v} = -\dfrac{1}{2} \ln y - C$$−v1​=−21​lny−C

Substituting $v = \dfrac{x}{y}$v=yx​:

$$\dfrac{y}{x} = \dfrac{1}{2} \ln y + C$$xy​=21​lny+C

Given $x(e) = e$x(e)=e, substituting $x = e$x=e and $y = e$y=e:

$$\dfrac{e}{e} = \dfrac{1}{2} \ln e + C$$ee​=21​lne+C

$$1 = \dfrac{1}{2} + C \Rightarrow C = \dfrac{1}{2}$$1=21​+C⇒C=21​

The particular solution is:

$$\dfrac{y}{x} = \dfrac{1}{2} \ln y + \dfrac{1}{2} = \dfrac{\ln y + 1}{2}$$xy​=21​lny+21​=2lny+1​

$$x = \dfrac{2y}{\ln y + 1}$$x=lny+12y​

To find $x(e^2)$x(e2), substitute $y = e^2$y=e2:

$$x(e^2) = \dfrac{2e^2}{\ln(e^2) + 1} = \dfrac{2e^2}{2 + 1} = \dfrac{2}{3} e^2$$x(e2)=ln(e2)+12e2​=2+12e2​=32​e2

Answer: $\dfrac{2}{3} e^2$32​e2