JEE Main 2023 — Differential Equations Question with Solution

Given,

$$\left(x^2-3y^2\right)dx+3xydy=0$$

$$\Rightarrow x^{2}dx-3y^{2}dx+3xydy=0$$

$$\Rightarrow \frac{x^{2}dx}{3xdx}-\frac{3y^{2}dx}{3xdx}+\frac{3ydy}{3xdx}=0$$

$$\Rightarrow \frac{x}{3}-\frac{y^2}{x}+y\frac{dy}{dx}=0$$

Let $$y^2=t$$

$$2y\frac{dy}{dx}=\frac{dt}{dx}$$

$$\therefore$$ $$\frac{1}{2}\frac{dt}{dx}-\frac{t}{x}+\frac{x}{3}=0$$

$$\Rightarrow \frac{dt}{dx}-\frac{2t}{x}=-\frac{2x}{3}$$

This is linear Differential Equation.

$$\text{ I.F. }=e^{\int -\frac{2}{x}dx}=e^{-2\ln |x|}=e^{\ln \frac{1}{x^2}}=\frac{1}{x^2}$$

So, $t \cdot \frac{1}{x^2}=\int \frac{1}{x^2}\left(\frac{-2 x}{3}\right) d x $

$\Rightarrow \frac{y^2}{x^2}=\frac{-2}{3}\ln |x|+C$

When, $x=1,y=1$

$$1=\frac{-2}{3}\ln (1)+C\Rightarrow C=1$$

$$\therefore \frac{y^2}{x^2}=\frac{-2}{3}\ln |x|+1$$

At $x=e,\frac{y^2(e)}{e^2}=\frac{-2}{3}+1$

$\Rightarrow y^2(e)=\frac{e^2}{3}$

$\Rightarrow 6y^2(e)=2e^2$