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 :

Answer 1

$y(1+xy)dx=xdy$

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

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

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

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

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

$\int xdx=\int \frac{1}{v^2}dv$

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

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

Put (1,-1)

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

we have to find $f(-\frac{1}{2})$

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

$\frac{{(-\frac{1}{2})}^2}{2}=\frac{-(-\frac{1}{2})}{y}-\frac{1}{2}$

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

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