Question

The solution of $(1-x^2)\frac{dy}{dx}+xy=x{y}^2$ is:

• A. $y(c+\sqrt{1-x^2})=1-x^2$
• B. $y(c-\sqrt{1-x^2})=1-x^2$
• C. $y(c+\sqrt{1-x^2})=\sqrt{1-x^2}$
• D. $y(c-\sqrt{1-x^2})=\sqrt{1-x^2}$

Answer 1

Answer: (C)

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

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

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

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

$\Rightarrow \int (\frac{1}{y-1}-\frac{1}{y})dy=-\frac{1}{2}\int \frac{-2x}{1-x^2}dx+c_1$

$\Rightarrow \log \left(\frac{y-1}{y}\right)=-\frac{1}{2}\log (1-x^2)+\log c_1$

$\Rightarrow \frac{y-1}{y}=\frac{c_1}{\sqrt{1-x^2}}$

$\Rightarrow y\sqrt{1-x^2}-\sqrt{1-x^2}=c_{1}y$

$\Rightarrow y(\sqrt{1-x^2}-c_1)=\sqrt{1-x^2}$

Put $c_1=-c$

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