Question

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

Answer 1

Answer: 8

$xdy-\left(y^2-4y\right)dx=0,x>0$
$\int \frac{dy}{y^2-4y}=\int \frac{dx}{x}$
$\int \left(\frac{1}{y-4}-\frac{1}{y}\right)dy=4\int \frac{dx}{x}$
${\log }_c|y-4|-{\log }_c|y|=4{\log }_{e}x+{\log }_{e}c$
$\frac{|y-4|}{|y|}=c{x}^4\stackrel{(1,2)}{\to }c=1$
$|y-4|=|y|x^4$
$C-1$
$y-4=y{x}^4$
$y=\frac{4}{1-x^4}$
$y(1)=\text{ ND (rejected) }$
$C-2$
$y-4=-y{x}^4$
$y=\frac{4}{1+x^4}$
$y(1)=2$
$y(\sqrt{2})=\frac{4}{5}\Rightarrow 10y(\sqrt{2})=8$