Question

If $y=f(x)(y>0)$ satisfies the differential equation $\frac{xdy}{dx}-y=\frac{x^4}{y}$ such that $y(1)=2$. Find the value of $y(-1)$.

Answer 1

Answer: 2

$xdy-ydx=\frac{x^4}{y}dx$
$\frac{y}{x}\cdot \frac{xdy-ydx}{x^2}=xdx$
$\frac{y}{x}d\left(\frac{y}{x}\right)=xdx\Rightarrow \frac{y^2}{x^2}=x^2+2c$
$\Rightarrow y=\sqrt{x^4+2x^2}\Rightarrow 4=1+2c\Rightarrow 2c=3$
$\therefore y=\sqrt{x^4+3x^2}\Rightarrow y(-1)=2$