Let $y=f(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^6+4x}{\sqrt{1-x^2}}$, $-1 < x < 1$ such that $f(0)=0$. If $6\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x)dx = 2\pi - \alpha$, then $\alpha^2$ is equal to
Solution:
Integrating Factor (I.F.) $= e^{\int \frac{x}{x^2-1} dx} = e^{\frac{1}{2} \ln|x^2-1|} = \sqrt{1-x^2}$.
Solution of the differential equation:
$$ y \cdot \sqrt{1-x^2} = \int \frac{x^6+4x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx $$
$$ y\sqrt{1-x^2} = \int (x^6+4x) dx = \frac{x^7}{7} + 2x^2 + C $$
Given $f(0)=0 \Rightarrow 0 = 0 + 0 + C \Rightarrow C=0$.
So, $f(x) = \frac{\frac{x^7}{7} + 2x^2}{\sqrt{1-x^2}}$.
We need to evaluate $I = 6\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x)dx$:
$$ I = 6 \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( \frac{\frac{x^7}{7}}{\sqrt{1-x^2}} + \frac{2x^2}{\sqrt{1-x^2}} \right) dx $$
The first term $\frac{x^7/7}{\sqrt{1-x^2}}$ is an odd function, so its integral is 0.
The second term is an even function, so:
$$ I = 6 \times 2 \int_{0}^{\frac{1}{2}} \frac{2x^2}{\sqrt{1-x^2}} dx = 24 \int_{0}^{\frac{1}{2}} \frac{x^2}{\sqrt{1-x^2}} dx $$
Substitute $x = \sin\theta, dx = \cos\theta d\theta$. Limits: $0 \to \frac{\pi}{6}$.
$$ I = 24 \int_{0}^{\frac{\pi}{6}} \frac{\sin^2\theta}{\cos\theta} \cos\theta d\theta = 24 \int_{0}^{\frac{\pi}{6}} \sin^2\theta d\theta $$
$$ = 24 \int_{0}^{\frac{\pi}{6}} \frac{1-\cos2\theta}{2} d\theta = 12 \left[ \theta – \frac{\sin2\theta}{2} \right]_{0}^{\frac{\pi}{6}} $$
$$ = 12 \left( \frac{\pi}{6} – \frac{\sin(\pi/3)}{2} \right) = 12 \left( \frac{\pi}{6} – \frac{\sqrt{3}}{4} \right) $$
$$ = 2\pi – 3\sqrt{3} $$
Comparing with $2\pi – \alpha$, we get $\alpha = 3\sqrt{3}$.
$$ \alpha^2 = (3\sqrt{3})^2 = 27 $$
Ans. 27