Integrate $t^2y'=-5ty$ I'm reading the notes and it says:
$t^2y'=-5ty$ where $y=y(t)$
Using direct integration we can conclude that $y=kt^{-5}$
I don't understand how this can be integrated directly. I can see that the solution is correct, but how did they achieve it?

Cancel $t$ at first sight. Conveniently separate the $x,y$ variables on either side of the equation.
$$\frac{dy}{dt}=-\frac{5 y }{t} $$
$$\frac{dy}{y} + \frac{5 \,dt }{t} = 0 $$
Both are logs on integration.
$$ y t ^5 = const. $$
