Hello,

I need to solve this separable differential equation

$\displaystyle 4y^3\frac{dy}{dx}=x^{-2}~~~~~,y(1) = 1$

We have $\displaystyle 4y^3~dy = \frac{1}{x^2}dx$

$\displaystyle y^4 = -\frac{1}{x}$

But what is $\displaystyle \bigg(-\frac{1}{x}\bigg)^{1/4} $?