Solve DE:y' + 2ty = 5t
The Attempt at a Solution
For yh: y' + 2ty = 0
y' = -2ty
Thus int(y-1,y) = int(-2t,t)
ln(y) = -t2 + C1
y = e-t2 + C1 = eC1e-t2 = Ce-t2
To find yp there are two ways to do it. With an integrating factor exp(-A(t)) where A'(t) = 2t or by saying yp = v(t)yh(t). Both ways will result in int(e-t2,t) which is unsolvable.
So I checked up the answer which read 5/2 + Cet2, so yp(t)=5/2. So I checked it by entering 5/2 in the DE and of course it was right.
Now my question is, how would you arrive at that answer? When would you make the assumption yp' = 0 and check if it is right? Isn't that too trivial? And when you would find an answer for yp' = 0, would you stop searching for another answer?