Question

Find the solution of the differential equation that satisfies the given initial condition of 

Answer 1

$x^2y'=y\sqrt{25-x^2}$

This ODE is separable. You can write

$\dfrac{\mathrm dy}y=\dfrac{\sqrt{25-x^2}}{x^2}\,\mathrm dx$

Integrating both sides gives

$\ln|y|=-\dfrac{\sqrt{25-x^2}}x-\arcsin\dfrac x5+C$

Given that $y(1)=1$, you find

$\ln1=-\sqrt{24}-\arcsin\dfrac15+C$
$\implies C=\sqrt{24}+\arcsin\dfrac15$