Question

Find the particular solutions of the following ODE:

y''+2y'-3y=e^t

Answer 1

\[y''+2y'-3y = e ^{t}\]

Answer 2

let us see your solution

Answer 3

\[y = e ^{t}\] ?

Answer 4

\[
e^t+2e^t-3e^t = 0\neq e^t
\]

Answer 5

You may want to try: \[
Ae^{Bt}
\]

Answer 6

\[
y = Ae^{Bt}\\
y' = BAe^{Bt}\\
y'' = B^2Ae^{Bt}\\
\]

Answer 7

Actually it's probably just \(y=Ae^t\) given how we know \(B=1\).

Answer 8

you have to multiply by x since r is positive 1

Answer 9

I suppose it could be \(Ae^t+Bte^t\)

Answer 10

well i believe you will have
Ae^-3t+Be^t+ z(t) and then you need to solve for the particular solution.

Answer 11

and when you solve for z(t) you will need to multiply by t, because you already have a e^t in the general solution.

Answer 12

It's actually y=Ate^t.
If I had y=e^t I would get 0=0.
Thanks though!