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## Homework Statement

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) = -t

^{2}+ C

_{1}

y = e

^{-t2 + C1}= e

^{C1}e

^{-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 + Ce

^{t2}, 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?

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