Question

A solution of the differential equation ${\left(\frac{dy}{dx}\right)}^2-x\frac{dx}{dx}+y=0isy=mx-4$ then the positive value of m is ___

Answer 1

$Y=mx-4\Rightarrow \frac{dy}{dx}=m{\left(\frac{dy}{dx}\right)}^2-x\frac{dy}{dx}+y=0\Rightarrow m^2-mx+mx-4=0\Rightarrow m^2=4\Rightarrow m=\pm 2$