Question

The solution of the differential equation $\left(1-x^2\right)\frac{dy}{dx}-xy=1$ is (where, $\left|x\right|<1,x\in R$ and $C$ is an arbitrary constant)

• A. $y\left(1-x^2\right)={\tan }^{-1}x+C$
• B. $y\sqrt{1-x^2}={\tan }^{-1}x+C$
• C. $y\sqrt{1-x^2}={\sin }^{-1}\left(x\right)+C$
• D. $y\cdot \left(1-x^2\right)={\sin }^{-1}x+C$

Answer 1

Answer: (C)

$\frac{dy}{dx}-\frac{x}{1-x^2}y=\frac{1}{1-x^2}$
L. $F.=e^{-\int \frac{x}{1-x^2}dx}$
$=e^{\frac{1}{2}\int -\frac{2x}{1-x^2}dx}=e^{\frac{1}{2}\ln \left(1-x^2\right)}=\sqrt{1-x^2}$
Hence, the solution of the differential equation is
$y\sqrt{1-x^2}=\int \frac{\sqrt{1-x^2}}{1-x^2}dx$
$y\sqrt{1-x^2}=\int \frac{1}{\sqrt{1-x^2}}dx$
$y\sqrt{1-x^2}={\sin }^{-1}(x)+C$