Question

If $y(x)$ is the solution of the differential equation
$x d y-\left(y^2-4 y\right) d x=0 \text { for } x>0, y(1)=2,$
and the slope of the curve $y=y(x)$ is never zero, then the value of $10 y(\sqrt{2})$ is ______

Answer 1

$ x d y-\left(y^2-4 y\right) d x=0, x>0$
$ \int \frac{d y}{y^2-4 y}=\int \frac{d x}{x} $
$\int\left(\frac{1}{y-4}-\frac{1}{y}\right) d y=4 \int \frac{d x}{x}$
$ \log _c|y-4|-\log _c|y|=4 \log _e x+\log _e c$
$ \frac{| y -4|}{| y |}= cx ^4 \xrightarrow{(1,2)} c =1 $
$|y-4|=|y| x^4 $
$ C -1 $
$ y-4=y x^4$
$ y=\frac{4}{1-x^4}$
$y(1)=\text { ND (rejected) } $
$C-2$
$ y-4=-y x^4 $
$ y=\frac{4}{1+x^4} $
$ y(1)=2$
$ y (\sqrt{2})=\frac{4}{5} \Rightarrow 10 y (\sqrt{2})=8 $