Question

If a curve $y=f\left(x\right)$ passes through the point $(1.-1)$ and satisfies the differential equation, $y(1+xy)dx=xdy$, then $f(-\frac{1}{2})$ is equal to

• A. $-\frac{4}{5}$
• B. $\frac{2}{5}$
• C. $\frac{4}{5}$
• D. $-\frac{2}{5}$

Answer 1

The correct option is C $\frac{4}{5}$

$\frac{y}{x}(1+xy)=\frac{dy}{dx}$

$y=vxv=\frac{y}{x}$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

$v(1+v^2{x}^2)=v+x\frac{dv}{dx}$

$x\frac{dv}{dx}=v^2{x}^2$

$\int \frac{dv}{v^2}=\int xdx$

$-\frac{1}{v}+c=\frac{x^2}{2}$

co-ordinates of point $(1,-1)$

$\therefore \frac{x^2}{2}=-\frac{x}{y}+c$

$\frac{1}{2}=\frac{-x}{y}-\frac{1}{2}$

$f(-\frac{1}{2}),x=\frac{-1}{2}$

$\frac{{\left(\frac{1}{2}\right)}^2}{2}=-\frac{(-v/2)}{y}=\frac{1}{2}$

$\Rightarrow \frac{1}{8}=\frac{1}{2y}-\frac{1}{2}$

$y=\frac{4}{5}$

$\therefore$ option $C$ is correct.