Solve the differential equation $$(4+t^2) \frac{dy}{dt} + 2ty = 4t$$
Solution:
$$(4+t^2) \frac{dy}{dt} + 2ty = \frac{d}{dt}[(4+t^2)y]$$
How?
Here's what I did:
$$\frac{d}{dt}[(4+t^2)y] = 4t$$
Then we integrate both side:
$$(4+t^2)y = 2t^2 + c$$
$$y = \frac{2t^2+c}{4+t^2}$$
I don't get the first step
On
If you don't know the trick, solve the homogeneous equation
$$(t^2+4)y'+2ty=0,$$ which is separable.
$$\frac{y'}y+\frac{2t}{t^2+4}=0$$
is solved by integration,
$$\log y+\log(t^2+4)=\log(y(t^2+4))=c,$$ and $$y(t^2+4)=c.$$
Now, by variation of the constant,
$$y(t^2+4)=c(t),$$ $$y'(t^2+4)+2ty=c'(t)=4t,$$ and finally
$$y=\frac{2t^2+c}{t^2+4}.$$
They use $(fg)'=f'g+g'f$. In this case $f=y(t)$ and $g=4+t^2$