Let y=f(x) be the solution of the differential equation (dy)/(dx)+xyx^2-1=x^6+4x√1-x^2,-1

Question

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 }_{1/2}^{-1/2}f(x)dx=2\pi -\alpha $ then ${\alpha }^{2}$ is equal to \_\_\_\_\_\_\_\_\_ .

Answer: 27

Accepted Answer

I.F. ${e}^{-\frac{1}{2}\int \frac{2x}{1-{x}^{2}}dx}={e}^{-\frac{1}{2}\ln (1-{x}^{2})}=\sqrt{1-{x}^{2}}$

$y	imes \sqrt{1-{x}^{2}}=\int ({x}^{6}+4x)dx=\frac{{x}^{7}}{7}+2{x}^{2}+c$

Given $y(0)=0\Rightarrow c=0$

$y=\frac{\frac{{x}^{7}}{7}+2{x}^{2}}{\sqrt{1-{x}^{2}}}$

$\begin{gathered} 6{\int }_{\frac{1}{2}}^{-\frac{1}{2}}\frac{\frac{{x}^{7}}{7}+2{x}^{2}}{\sqrt{1-{x}^{2}}}dx=6{\int }_{\frac{1}{2}}^{-\frac{1}{2}}\frac{2{x}^{2}}{\sqrt{1-{x}^{2}}}dx \ =24{\int }_{\frac{1}{2}}^{0}\frac{{x}^{2}}{\sqrt{1-{x}^{2}}}dx \end{gathered}$

$\begin{gathered} x=\sin 	heta \ dx=\cos 	heta d	heta \ =24{\int }_{\frac{\pi }{6}}^{0}\frac{{\sin }^{2}	heta }{\cos 	heta }\cos 	heta d	heta \ =24{\int }_{\frac{\pi }{6}}^{0}(\frac{1-\cos 2	heta }{2})d	heta =12{[	heta -\frac{\sin 2	heta }{2}]}_{\frac{\pi }{6}}^{0} \ =12(\frac{\pi }{6}-\frac{\sqrt{3}}{4}) \ =2\pi -3\sqrt{3} \ {\alpha }^{2}=(3\sqrt{3}{)}^{2}=27 \end{gathered}$

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