JEE Main 2025 Apr 7 Shift 2, Mathematics Q13: Methods of Solving First Order, First Degree Differential Equation

Since $x^4+2x^2+1=(x^2+1{)}^2$, dividing through by $(x^2+1)$ gives the linear equation $\frac{dy}{dx}-\frac{2x}{x^2+1}y=(x^2+1)\cos x$.

The integrating factor is $e^{-\int \frac{2x}{x^2+1}dx}=\frac{1}{x^2+1}$, so $\frac{d}{dx}\left[\frac{y}{x^2+1}\right]=\cos x$.

Integrating: $\frac{y}{x^2+1}=\sin x+C$. Using $y(0)=1$: $1=0+C\Rightarrow C=1$, so $y=(x^2+1)(\sin x+1)$.

${\int }_{-3}^{3}ydx={\int }_{-3}^3(x^2+1)\sin xdx+{\int }_{-3}^3(x^2+1)dx$. The first integrand is odd, so it integrates to 0 over the symmetric interval.

The second integral: ${\int }_{-3}^3(x^2+1)dx=2{\int }_0^3(x^2+1)dx=2{\left[\frac{x^3}{3}+x\right]}_0^3=2(9+3)=24$.

$\therefore {\int }_{-3}^{3}ydx=0+24=24$.