Let y=y(x) be the solution of the differential equation (1-x^2)dy=(xy+(x^3+2)√1-x^2)dx,-1<x<1 and y(0)=0. If ∫ _- 1 2^ 1 2√1-x^2y(x)dx=k then k^-1 is…

Given,

$$(1-{x}^{2})\frac{dy}{dx}=xy+({x}^{3}+2)\sqrt{1-{x}^{2}}$$(1−x2)dxdy​=xy+(x3+2)1−x2​

$$\Rightarrow \frac{dy}{dx}+(\frac{-x}{1-{x}^{2}})y=\frac{{x}^{3}+2}{\sqrt{1-{x}^{2}}}$$⇒dxdy​+(1−x2−x​)y=1−x2​x3+2​

$$\mathrm{IF}={e}^{\int \frac{-x}{1-{x}^{2}}dx}=\sqrt{1-{x}^{2}}$$IF=e∫1−x2−x​dx=1−x2​

$$y(x)\cdot \sqrt{1-{x}^{2}}=\frac{{x}^{4}}{4}+2x+c$$y(x)⋅1−x2​=4x4​+2x+c

$$y(0)=0\Rightarrow c=0$$y(0)=0⇒c=0

$$\sqrt{1-{x}^{2}}y(x)=\frac{{x}^{4}}{4}+2x$$1−x2​y(x)=4x4​+2x

So,required value $\displaystyle ={\int }_{-\frac{1}{2}}^{\frac{1}{2}}(\frac{{x}^{4}}{4}+2x)dx-\frac{1}{4}\cdot 2{\int }_{0}^{\frac{1}{2}}{x}^{4}dx$=∫−21​21​​(4x4​+2x)dx−41​⋅2∫021​​x4dx

$$=\frac{1}{10}{({x}^{5})}_{0}^{\frac{1}{2}}=\frac{1}{320}$$=101​(x5)021​​=3201​

$${k}^{-1}=320$$k−1=320