Question

The solution of the differential equation $\frac{d y}{d x}+\frac{x y}{1 - x^{2}}=x\sqrt{y},$ $\left(\left|x\right| < 1\right)$ is $\sqrt{y}=-\frac{f \left(x\right)}{3}+C\left(1 - x^{2}\right)^{\frac{1}{4}},$ where $f\left(\frac{1}{2}\right)=\frac{3}{4}$ and $C$ is an arbitrary constant. Then, the value of $f\left(- \frac{1}{2}\right)$ is

• A. $-\frac{3}{4}$
• B. $\frac{3}{4}$
• C. $\frac{1}{4}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

Given equation is $\frac{1}{\sqrt{y}}\frac{d y}{d x}+\frac{x}{1 - x^{2}}\sqrt{y}=x$
Let, $2\sqrt{y}=\nu$
$\Rightarrow \frac{1}{\sqrt{y}}\frac{d y}{d x}=\frac{d \nu}{d x}$
Thus, we have
$\frac{d \nu}{d x}+\frac{x}{2 \left(1 - x^{2}\right)}\nu=x$
$\therefore $ I.F. $=e^{\displaystyle \int \frac{x}{2 \left(1 - x^{2}\right)} d x}$
$=e^{\displaystyle \int \frac{- d \left(1 - x^{2}\right)}{4 \left(1 - x^{2}\right)}}$
$=e^{- \frac{1}{4} ln \left(1 - x^{2}\right)}$
$=\left(1 - x^{2}\right)^{- \frac{1}{4}}$
Thus, the solution is $\nu\left(1 - x^{2}\right)^{- \frac{1}{4}}=\displaystyle \int x \left(1 - x^{2}\right)^{- \frac{1}{4}}dx$
or $\nu\cdot \left(1 - x^{2}\right)^{- \frac{1}{4}}=-\frac{2}{3}\left(1 - x^{2}\right)^{\frac{3}{4}}+C^{'}$
or $2\sqrt{y}=-\frac{2}{3}\left(1 - x^{2}\right)+C^{'}\left(1 - x^{2}\right)^{\frac{1}{4}}$
$\Rightarrow \sqrt{y}=-\frac{\left(1 - x^{2}\right)}{3}+C\left(1 - x^{2}\right)^{\frac{1}{4}}$
$\Rightarrow f\left(x\right)=1-x^{2}$
$\Rightarrow f\left(- \frac{1}{2}\right)=1-\frac{1}{4}=\frac{3}{4}$