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?

There's no need to write $y'=\frac {dy}{dt}$: you just need to know that $\frac {y'}y=(\ln y)'$. Then
$$\frac {y'}y=-5t^{-1}$$
Integrating with respect to $dt$ (without multiplying or doing any differential-shenanigans), we get
$$
\ln y =-5\ln t+c
$$
And the result follows :-).
