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A tricky 2nd Oder ODE Problem,nonhomo delta=0 case

  1. May 11, 2009 #1
    solve the initial value problem:
    y''+8.4y'+17.64y=e-4.2x
    y(0)=1, y'(0)=1

    y(x)=?

    The way I tried to achieve is to solve the corresponding homo equation first:
    y''+8.4y'+17.64y=0, which gives yc;
    yc=c1e-4.2x+c2e-4.2xx

    Then try to find yp, generally I would assume a yp=Ae-4.2x, but from the yc got above, clearly yp=Ae-4.2x or yp=Ae-4.2xx is not good. If I add one more 'x' in yp assumption,in which it seems A has to be 0, which is not right either....

    Any idea? Or I made some mistake?

    Meric.
     
  2. jcsd
  3. May 11, 2009 #2

    HallsofIvy

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    Because both [itex]e^{-4.2x}[/itex] and [itex]xe^{-4.2x}[/itex] satisfy the homogeneous equation, you need "go up" one more x: try
    [tex]y_p= x^2e^{-4.2x}[/tex]

    Was that what you meant by "add one more 'x' in yp assumption"?

    If
    [tex]y_p= Ax^2e^{-4.2x}[/tex]
    then
    [tex]y_p'= 2Axe^{-4.2x}- 4.2Ax^2e^{-4.2x}[/tex]
    and
    [tex]y_p"= 2Ae^{-4.2x}- 16.8Axe^{-4.2x}+ 17.64Ax^2e^{-4.2x}[/tex]
    Putting that into the differential equation, all the terms involving [itex]xe^{-4.2x}[/itex] or [itex]x^2e^{-4.2x}[/itex] will cancel leaving only [itex]2Ae^{-4.2}= e^{-4.2x}[/itex] so A is NOT 0.
     
    Last edited: May 11, 2009
  4. May 11, 2009 #3
    Oh I made a silly mistake in differentiate the Yp...

    Yeah, you are right, I got the right answer now. Thanks
     
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