1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find General Solution to Equation

  1. May 26, 2012 #1
    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.
     
    Last edited: May 26, 2012
  2. jcsd
  3. May 26, 2012 #2

    sharks

    User Avatar
    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.
     
  4. May 26, 2012 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Correct so far.


    What are C and D? Show your work in detail, please.

    ehild
     
  5. May 26, 2012 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Subbing into the top equation, I get e2x((-C+D)cosx + (-C-D)sinx) for the left hand side.
     
  6. May 26, 2012 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    That is correct, Sammy. :cool:

    ehild
     
  7. May 26, 2012 #6
    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.
     
  8. May 26, 2012 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What do you get for y' and y'' ?
     
  9. May 27, 2012 #8
    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.
     
  10. May 27, 2012 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    To repeat my question(s) ...

    What do you get for yP' ?

    What do you get for yP'' ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook