JEE Main 2025 Apr 7 Shift 1, Mathematics Q6: Methods of Solving First Order, First Degree Differential Equation

Step 1: Cast as a linear ODE.

$x(x^2+e^x)\frac{dy}{dx}+e^x(x-2)y=x^3$

$\frac{dy}{dx}+\frac{e^x(x-2)}{x(x^2+e^x)}y=\frac{x^2}{x^2+e^x}$

Step 2: Find the integrating factor. Notice that

$\frac{d}{dx}\ln \left(1+\frac{e^x}{x^2}\right)=\frac{e^x(x-2)/x^3}{1+e^x/x^2}=\frac{e^x(x-2)}{x(x^2+e^x)}$

So the coefficient of $y$ is exactly $\frac{d}{dx}\ln \left(1+\frac{e^x}{x^2}\right)$, giving

$\text{I.F.}=1+\frac{e^x}{x^2}$

Step 3: Multiply through and integrate.

$\frac{d}{dx}\left[y\left(1+\frac{e^x}{x^2}\right)\right]=\frac{x^2}{x^2+e^x}\cdot \frac{x^2+e^x}{x^2}=1$

Integrating: $y\left(1+\frac{e^x}{x^2}\right)=x+C$.

Step 4: Apply the initial condition $(1,0)$. $0=1+C\Rightarrow C=-1$.

$y=\frac{x-1}{1+e^x/x^2}$

Step 5: Evaluate at $x=2$.

$y(2)=\frac{1}{1+e^2/4}=\frac{4}{4+e^2}$

$\therefore$ Option (4).