## Find General Solution to Equation

1. The problem statement, all variables and given/known data

y'' - 3y' + 2y = e2xcos x

3. The attempt at a solution

So this is a second order inhomogeneous equation.

gives λ = {2,1}
get yCF giving Ae2x + Bex
so yP = e2x(Ccosx + Dsinx)

Here is where is gets a bit fuzzy.

Subbing this into the top equation gives me 2e2x((-2D-C)cosx + (2C-D)sinx)

which is wrong, i have gone through it and am missing something can you help me find this correct expression? Thanks.

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 Recognitions: Gold Member Well, it depends on what method you used to find ##y_p##. Personally, i use the inverted operator technique (but there is apparently also the variation of parameters method). Here is what i would do: Use the shift theorem to separate ##e^{2x}## (moves to the left) from ##\cos x##. $$\frac{ke^{ax}.\phi (x)}{L(D)}\to ke^{ax}.\frac{\phi (x)}{L(D+a)}$$where k=1, a=2. From there, use the appropriate inverted operator technique on ##\cos x## which is: $$\frac{k\cos (ax+b)}{L(D)} \to \frac{k\cos (ax+b)}{L(-a^2)}$$ where k=1, a=1, b =0.

Recognitions:
Homework Help
 Quote by JoshMaths So this is a second order inhomogeneous equation. gives λ = {2,1} get yCF giving Ae2x + Bex so yP = e2x(Acosx + Bsinx)
Correct so far.

 Quote by JoshMaths Here is where is gets a bit fuzzy. Subbing this into the top equation gives me 2e2x((-2D-C)cosx + (2C-D)sinx)

ehild

Mentor

## Find General Solution to Equation

 Quote by JoshMaths 1. The problem statement, all variables and given/known data y'' - 3y' + 2y = e2xcos x 3. The attempt at a solution So this is a second order inhomogeneous equation. gives λ = {2,1} get yCF giving Ae2x + Bex so yP = e2x(Ccosx + Dsinx) Here is where is gets a bit fuzzy. Subbing this into the top equation gives me 2e2x((-2D-C)cosx + (2C-D)sinx) which is wrong, i have gone through it and am missing something can you help me find this correct expression? Thanks.
Subbing into the top equation, I get e2x((-C+D)cosx + (-C-D)sinx) for the left hand side.

Recognitions:
Homework Help
 Quote by SammyS Subbing into the top equation, I get e2x((-C+D)cosx + (-C-D)sinx) for the left hand side.
That is correct, Sammy.

ehild

 Quote by ehild Correct so far. What are C and D? Show your work in detail, please. ehild
Yep, my mistake, C and D were in y particular but I always mix them up accidentally. It has been fixed.

 Quote by SammyS Subbing into the top equation, I get e2x((-C+D)cosx + (-C-D)sinx) for the left hand side.
Yep, this is right, i tried and i got (-C-3D)cosx +(3C-D)sinx i guess i am just not seeing it. I understand if you don't have time to write out some of the algebra manipulations but they would be useful if anyone could find a minute. If not, i'll keep working at it.

Mentor
 Quote by JoshMaths Yep, my mistake, C and D were in y particular but I always mix them up accidentally. It has been fixed. Yep, this is right, i tried and i got (-C-3D)cosx +(3C-D)sinx i guess i am just not seeing it. I understand if you don't have time to write out some of the algebra manipulations but they would be useful if anyone could find a minute. If not, i'll keep working at it.
What do you get for y' and y'' ?

 Quote by SammyS What do you get for y' and y'' ?
I have tried with y'' + 3y' + 2y = e2xcos x

and get e2x((11C+7D)cosx + (11D-7C)sinx) which is correct so i think it was just that specific example i was having trouble with, but i understand the principle behind it better now.

 Mentor To repeat my question(s) ... What do you get for yP' ? What do you get for yP'' ?

 Tags calculus, inhomogeneous, particular solution, solve