Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with Complex Second Order Equations

  1. Feb 13, 2005 #1
    Ok, this one is really sticking me up:

    [tex]y'' - 2y' + 2y = e^{t}cos(t)[/tex]

    I solved the homogenous version and got roots 1 +/- i and put these in to get the equation

    [tex]y_h = c_1e^tcos(t) + c_1e^tsin(t)[/tex]

    And I found that the root for e^tcos(t) should be (D- (1 +/- i)

    But I'm completely stuck on what to do after this, the part that really confuses me is that in the answer, they only have a e^t sin(t) factor in the answer in addition to the normal homogenous answer, why does sin(t) appear without cos(t) ?

    Thanks for your help,

  2. jcsd
  3. Feb 13, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    Did u try to apply Lagrange's method...?

  4. Feb 13, 2005 #3

    I don't believe so, I don't think we were ever taught about that, is there another way to solve it? Maybe we called it something different. I'll go back and look through my notes

  5. Feb 13, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Search for "Method of Variation of Constants" or "Method of Lagrange"...

  6. Feb 13, 2005 #5
    ok, I got it, thanks
  7. Feb 15, 2005 #6


    User Avatar
    Science Advisor

    You should be able to do this by basic "undetermined coefficients".

    Normally, if you have "right hand side" like etcos(t), you would try
    y= et(Acos(t)+ Bsin(t)).

    However, here, functions of that form already satisfy the homogenous equation and can't give you anything but 0 on the right side.

    Okay, so you multiply by t:
    try y= ett(Acos(t)+ Bsin(t)).

    The rest is cranking away.
  8. Feb 15, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    You guys mind if I finish up? He's gone to something else I think. You know school and all:

    [tex]y'' - 2y' + 2y = e^{t}cos(t)[/tex]

    Because the RHS is a particular solution of a linear ODE, namely:

    [tex] (D^2-2D+2)y=0 [/tex]

    can apply this operator to both side of non-homogeneous eq. to collapse the RHS:

    [tex] (D^2-2D+2)(D^2-2D+2)y=0 [/tex]

    Since the solution to this equation is:

    [tex]y=c_1e^x\cos(x)+c_2e^x\sin(x)+Axe^x\sin(x)+Bxe^x\cos(x) [/itex]

    then the particular solution to the non-homogeneous equation must be of the form:

    [tex]y_p=Axe^x\sin(x)+Bxe^x\cos(x) [/tex]

    Substituting this equation into the non-homogeneous equation and equating coefficients, we find B=0 and A=1/2.

    Thus the general solution is:

    [tex] y(x)=c_1e^x\cos(x)+c_2e^x\sin(x)+\frac{1}{2}xe^x\sin(x)[/tex]

    Using [itex]c_1=1[/itex] and [itex]c_2=0[/itex], I plotted a graph of the solution which is attached.

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook