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Solving second order non homogeneous differential equation

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is to solve:

    [itex]y''-2y'+5y = e^{x}(cos^{2}(x)+x^{2})[/itex]


    2. Relevant equations

    3. The attempt at a solution

    I (think I) have solved the associated homogeneous equation:

    [itex]y''-2y'+5y = 0[/itex]

    giving the solution as:

    [itex]y_{h} = e^{x}(C_{1}cos(2x)+C_{2}sin(2x))[/itex]
    (This could of course be rearranged using the trig identities.... would that help?)

    And now I don't know how to go about finding a particular solution, could somebody please point me in the right direction as to which method I could use?
    Thanks a lot!
     
  2. jcsd
  3. Apr 22, 2012 #2

    LCKurtz

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    I would suggest the method of finding a particular solution with the method of Undetermined Coefficients, more formally known as the method of annihilators. Have you studied that?
     
  4. Apr 22, 2012 #3
    Yes I have studied this method (a little) however I am struggling to choose the right form for the particular solution:

    Should [itex]y_{p} = Asin(2x)+Bcos(2x)+e^{x}(C+Dx+Ex^{2})[/itex] work?
     
  5. Apr 22, 2012 #4

    LCKurtz

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    That ##\cos^2x## on the right side of your DE is ##\frac{1+\cos(2x)} 2## so you basically have a term ##e^x(\frac{1+\cos(2x)} 2)## in the NH term. So there's no reason to expect ##\sin(2x)## and ##\cos(2x)## to work. You would want to multiply them by ##e^x##. But that is in the complementary solution. So how do you fix that?
     
  6. Apr 22, 2012 #5
    would simply multiplying by [itex]x[/itex] solve this?
    i.e.
    [itex]y_{p} = e^{x}(Axsin(2x)+Bxcos(2x)+C+Dx+Ex^{2})[/itex]
     
  7. Apr 22, 2012 #6

    LCKurtz

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    Well, that's the standard fix for that situation. I think that is a good trial for the particular solution. Only way to be sure you haven't overlooked something is to go ahead and try it
     
  8. Apr 22, 2012 #7

    sharks

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    How about the method of inverted operator techniques? (which i personally find easier). Although some people use the Wronskian method, i think.

    The D operator is simply [itex]\frac{d}{dx}[/itex].
    Expressing first, the L.H.S. in terms of D: [itex]L(D)=D^2-2D+5[/itex], where L(D) is a function of D.
    [tex]L(D)y_p=e^x(cos^{2}(x)+x^{2})[/tex]
    So, [tex]y_p=\frac{1}{L(D)}(e^xcos^{2}(x)+e^xx^{2})
    \\y_p=\frac{1}{L(D)}e^xcos^{2}(x)+\frac{1}{L(D)}e^xx^{2}[/tex]
    Here, [itex]y_p=y_1+y_2[/itex],
    [tex]y_1=\frac{1}{L(D)}e^xcos^{2}(x)[/tex]
    [tex]y_2=\frac{1}{L(D)}e^xx^{2}[/tex]
    Now, you have to use the Shift theorem to evaluate both [itex]y_1[/itex] and [itex]y_2[/itex].
    The Shift theorem simply shifts D to (D+a) and simultaneously separates [itex]e^x[/itex] from [itex](cos^2x+x^2)[/itex].

    To find [itex]y_1[/itex]:
    [tex]y_1=\frac{1}{L(D)}e^xcos^{2}(x)=\frac{1}{(D^2-2D+5)}e^xcos^{2}(x)[/tex]
    Applying the Shift theorem:
    [tex]y_1=e^x\frac{1}{((D+1)^2-2(D+1)+5)}cos^{2}(x)
    \\=e^x\frac{1}{(D^2+5)}cos^{2}(x)
    \\=e^x\frac{1}{(D^2+5)}(\frac{1}{2}+\frac{1}{2} \cos 2x)
    \\=e^x\frac{1}{(D^2+5)}(\frac{1}{2})+e^x\frac{1}{(D^2+5)}(\frac{1}{2} \cos 2x)
    \\=e^x(\frac{1}{10})+e^x(\frac{1}{2} \cos 2x)
    \\=e^x(\frac{1}{10}+\frac{1}{2} \cos 2x)
    [/tex]
    Now, find [itex]y_2[/itex]. Applying the Shift theorem again, you'll get:
    [tex]y_2=e^x\frac{1}{(D^2+5)}x^2
    \\=e^x(\frac{1}{5}-\frac{D^2}{25})x^2
    \\=e^x(\frac{1}{5}x^2-\frac{2}{25})
    [/tex]
    Therefore, [tex]y_p=e^x(\frac{1}{10}+\frac{1}{2} \cos 2x)+e^x(\frac{1}{5}x^2-\frac{2}{25})
    \\=\frac{1}{50}e^x+\frac{1}{5}e^xx^2+ \frac {1}{2} e^x \cos 2x[/tex]
    The final answer (the general solution) is: [itex]y_c+y_p[/itex] where [itex]y_c[/itex] is the complementary function that you've already obtained in your first post.
     
    Last edited: Apr 22, 2012
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