Let $y=y(x)$ be the solution of the differential equation $\left(1-x^{2}\right) d y=\left(x y+\left(

Question

Let $y=y(x)$ be the solution of the differential equation $\left(1-x^{2}\right) d y=\left(x y+\left(x^{3}+2\right) \sqrt{1-x^{2}}\right) d x,-1<x<1$, and $y(0)=0$. If $\int_{\frac{-1}{2}}^{\frac{1}{2}} \sqrt{1-x^{2}} y(x) d x=k$, then $k^{-1}$ is equal to

Professor PC Thomas Classes · Answer

(320) : Given differential equation is\ $\left(1-x^{2}ight) d y=\left(x y+\left(x^{3}+2ight) \sqrt{1-x^{2}}ight) d x$,\ $-1<x<1$\ $\frac{d y}{d x}=\frac{x y+\left(x^{3}+2ight) \sqrt{1-x^{2}}}{1-x^{2}}$\ $\Rightarrow \quad \frac{d y}{d x}-\left(\frac{x}{1-x^{2}}ight) y=\frac{x^{3}+2}{\sqrt{1-x^{2}}}$\ I.F. $=e^{\int \frac{-x}{1-x^{2}} d x}=\sqrt{1-x^{2}}$\ $	herefore \quad$ The solution of the given D.E. is\ $y \cdot \sqrt{1-x^{2}}=\int\left(2+x^{3}ight) d x+c$\ $\Rightarrow y \sqrt{1-x^{2}}=2 x+\frac{x^{4}}{4}+c$\ When $x=0, y=0 	herefore 0=0+0+c \Rightarrow c=0$\ From (i) $y \sqrt{1-x^{2}}=2 x+\frac{x^{4}}{4}$\ Now, $\int_{-1 / 2}^{1 / 2}\left(2 x+\frac{x^{4}}{4}ight) d x=\left[x^{2}+\frac{x^{5}}{20}ight]_{-1 / 2}^{1 / 2}=\frac{1}{320}$\