I'm trying to review some Calculus for the year... I need help with this problem.?

Question

Find the general solution to:

2*t*y'+ 4*y = 3...

I feel so silly not even remembering where to start on something like this.

Answer 1

This is a separable equation. We have
2*t*y' = 3 - 4*y, so y'/(3 - 4*y) =1/(2*t) or
dy/(3 - 4*y) = dt/(2*t). Integrate to get (ln(3 - 4y))/(-4) =
(ln t)/2 + C.

Answer 2

This is really diff. equations rather than calculus, but whatever. So you're looking for a function y(t) such that
2ty' + 4y = 3 Because of the factor t in the first term you're going to have a t^2 in your answer to get rid of it. So try
y = at^2 +bt +c
This gives 2t(2at +b) +4(at^2 + bt +c) = 3
So: t^2 term: 4a +4a = 0 or a=0
t term: 2b + 4b = 0 or b=0
constant term: 4c =3 or c = 3/4
So y=3/4 which satisfies the d.e.
To be honest, I've forgotten a lot of my d.e. so I'm not sure if the function y = at^2 . . . could generally have been reduced. Obviously, in this case I over compensated.