Question

Let $y=y\left(x\right)$ be the solution of the differential equation
$(1-x^2)\frac{dy}{dx}-xy=1,xϵ(-1,1).$ if $y\left(o\right)=0,$ then $y\left(\frac{1}{2}\right)$ is equal to

• A. $\frac{\pi }{6\sqrt{3}}$
• B. $\frac{\pi }{\sqrt{3}}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (A)

$\frac{\pi }{6\sqrt{3}}$

$(1-x^2)\frac{dy}{dx}-xy=1$

$\frac{dy}{dx}-\frac{x}{1-x^2}y=\frac{1}{1-x^2}$

$\frac{dy}{dx}+P\left(x\right)g=Q\left(x\right)$

Solving linear differential equation by calculating $I.F$

$I.F=e^{\int P\left(x\right)dx}$

$=e^{\int -\frac{x}{1-x^2}dx}$

$=e^{\frac{1}{2}\int \frac{d(-x^2)}{1-x^2}}$

$=e^{1/2ln|1-x^2|}$

$=e^{ln{|1-x^2|}^{1/2}}$

$IF=\sqrt{1-x^2}$

So, $y\left(IF\right)=\int Q\left(x\right)\left(IF\right)dx$

$y\sqrt{1-x^2}=\int \frac{x}{1-x^2}\sqrt{1-x^2}dx$

$y\left(\sqrt{1-x^2}\right)=\int \frac{x}{\sqrt{1-x^2}}dx$

$y\left(\sqrt{1-x^2}\right)=\frac{1}{2}{\sin }^{-1}x+C$

$y\left(\sqrt{1-x^2}\right)=\frac{1}{2}{\sin }^{-1}x+C$

$y\left(0\right)=0$

$0=\frac{1}{2}{\sin }^{-1}0+C$

$C=0$

So, $y\left(\sqrt{1-x^2}\right)=\frac{1}{2}{\sin }^{-1}x$

At $x=\frac{1}{2}$

$y=\frac{1}{2}\frac{{\sin }^{-1}\left(\frac{1}{2}\right)}{\sqrt{1-{\left(\frac{1}{2}\right)}^2}}=\frac{1}{2}\times \frac{\pi }{6}\times \frac{2}{\sqrt{3}}=\frac{\pi }{6\sqrt{3}}$