JEE Main 2025 Jan 22 Shift 2, Mathematics Q21: Methods of Solving First Order, First Degree Differential Equation

For the linear ODE, the integrating factor is $\mu =e^{\int \frac{x}{x^2-1}dx}=e^{\frac{1}{2}\ln |x^2-1|}$. Since $|x|<1$, $x^2-1<0$, so $|x^2-1|=1-x^2$, giving $\mu =\sqrt{1-x^2}$.

Multiplying through:

$\frac{d}{dx}\left(y\sqrt{1-x^2}\right)=(x^6+4x)\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}=x^6+4x$.

Integrating: $y\sqrt{1-x^2}=\frac{x^7}{7}+2x^2+C$.

At $x=0$, $y=0$: $0=C$, so $C=0$ and $y=\frac{\frac{x^7}{7}+2x^2}{\sqrt{1-x^2}}$.

For ${\int }_{-1/2}^{1/2}$, the odd term $\frac{x^7/7}{\sqrt{1-x^2}}$ integrates to 0. So:

$6{\int }_{-1/2}^{1/2}\frac{2x^2}{\sqrt{1-x^2}}dx=24{\int }_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}dx$ (even integrand).

Substitute $x=\sin \varphi$, $dx=\cos \varphi d\varphi$:

$24{\int }_0^{\pi /6}\frac{{\sin }^2\varphi \cos \varphi }{\cos \varphi }d\varphi =24{\int }_0^{\pi /6}{\sin }^2\varphi d\varphi =24{\int }_0^{\pi /6}\frac{1-\cos 2\varphi }{2}d\varphi$

$=12{\left[\varphi -\frac{\sin 2\varphi }{2}\right]}_0^{\pi /6}=12\left(\frac{\pi }{6}-\frac{\sin (\pi /3)}{2}\right)=12\left(\frac{\pi }{6}-\frac{\sqrt{3}}{4}\right)=2\pi -3\sqrt{3}$.

Therefore $\alpha =3\sqrt{3}$, and ${\alpha }^2=27$.