Mathematics - Differential Equation Question with Solution | TestHub

I.F. ${\mathrm{e}}^{-\frac{1}{2}\int \frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx}}={\mathrm{e}}^{-\frac{1}{2}\ln \left(1-{\mathrm{x}}^2\right)}=\sqrt{1-{\mathrm{x}}^2}$ $y\times \sqrt{1-x^2}=\int \left(x^6+4x\right)dx=\frac{x^7}{7}+2x^2+c$ Given $y(0)=0\Rightarrow c=0$ $y=\frac{\frac{x^7}{7}+2x^2}{\sqrt{1-x^2}}$ Now, $6{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{\frac{x^7}{7}+2{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}=6{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\frac{2{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$ $\begin{array}{rl}& =24{\int }_0^{\frac{1}{2}}\frac{x^2}{\sqrt{1-x^2}}dx\\ & \text{ Put }x=\sin \theta \\ & dx=\cos \theta d\theta \\ & =24{\int }_0^{\frac{\pi }{6}}\frac{{\sin }^2\theta }{\cos \theta }\cos \theta d\theta \\ & =24{\int }_0^{\frac{\pi }{6}}\left(\frac{1-\cos 2\theta }{2}\right)d\theta =12{\left[\theta -\frac{\sin 2\theta }{2}\right]}_0^{\frac{\pi }{6}}\\ & =12\left(\frac{\pi }{6}-\frac{\sqrt{3}}{4}\right)\\ & =2\pi -3\sqrt{3}\\ & {\alpha }^2=(3\sqrt{3}{)}^2=27\end{array}$