Question

The solution of the differential equation $\left(1 - x^{2}\right)\frac{d y}{d x}-xy=1$ 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)

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