Question

Solve the differential equation: $(1-x^2)\frac{dy}{dx}-xy=\frac{1}{\sqrt{(1-x^2)}}$

• A. $y\sqrt{(1-x^2)}=\sin x+c.$
• B. $y\sqrt{(1+x^2)}={\sin }^{-1}x+c.$
• C. $y\sqrt{(1-x^2)}={\sin }^{-1}x+c.$
• D. None of these.

Answer 1

The correct option is C $y\sqrt{(1-x^2)}={\sin }^{-1}x+c.$

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

$\Rightarrow \frac{dy}{dx}-\frac{xy}{(1-x^2)}=\frac{1}{{(1-x^2)}^{3/2}}$ ...(1)
Here $P=-\frac{x}{(1-x^2)}\Rightarrow \int Pdx=-\int \frac{x}{(1-x^2)}dx$

$=\frac{1}{2}\log (1-x^2)=\log \sqrt{(1-x^2)}$
$\therefore I.F.=e^{\log \sqrt{(1-x^2)}}=\sqrt{(1-x^2)}$
Multiplying (1) by $I.F.$ we get
$\sqrt{(1-x^2)}\frac{dy}{dx}-\frac{xy}{\sqrt{(1-x^2)}}=\frac{1}{(1-x^2)}$
Integrating both side
$y\sqrt{(1-x^2)}=\int \frac{1}{(1-x^2)}dx+c$

$\Rightarrow y\sqrt{(1-x^2)}={\sin }^{-1}x+c$