Show that this piece-wise function defines a differentiable solution

Question

Show that

$y(x) = \begin{cases}-x^4 & x < 0, \\ x^4 & x \geqq 0 \end{cases}$

defines a differentiable solution of $xy'=4y$ for all $x$, but is not of the form $y(x)=Cx^4$.

Answer 1

Given your definition of y(x), we have that: $$y'(x):$$$$=-4x^3, x < 0$$ $$=4x^3, x \geq 0$$

substitute these values into $xy'=4y$

for $x<0$ you have $x(-4x^3)=4(-x^4)$ which reduces to $x^4=x^4$

you take the same steps to verify for $x>0$.