Question

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