sanchez9457
  • sanchez9457
Find the particular solutions of the following ODE: y''+2y'-3y=e^t
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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sanchez9457
  • sanchez9457
\[y''+2y'-3y = e ^{t}\]
nincompoop
  • nincompoop
let us see your solution
sanchez9457
  • sanchez9457
\[y = e ^{t}\] ?

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anonymous
  • anonymous
\[ e^t+2e^t-3e^t = 0\neq e^t \]
anonymous
  • anonymous
You may want to try: \[ Ae^{Bt} \]
anonymous
  • anonymous
\[ y = Ae^{Bt}\\ y' = BAe^{Bt}\\ y'' = B^2Ae^{Bt}\\ \]
anonymous
  • anonymous
Actually it's probably just \(y=Ae^t\) given how we know \(B=1\).
anonymous
  • anonymous
you have to multiply by x since r is positive 1
anonymous
  • anonymous
I suppose it could be \(Ae^t+Bte^t\)
anonymous
  • anonymous
well i believe you will have Ae^-3t+Be^t+ z(t) and then you need to solve for the particular solution.
anonymous
  • anonymous
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.
sanchez9457
  • sanchez9457
It's actually y=Ate^t. If I had y=e^t I would get 0=0. Thanks though!

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