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Constant Coefficient Differential Equation

  1. Aug 22, 2010 #1
    Hey there

    I am new to Physics forums and could use some help understanding/solving this problem.

    Use transformation x=e^t to convert equation
    x^2y'' + 10xy' + 8y = x^2

    Solve this equation to show that solution is

    y = a/x^8 + b/x + x^2/30

    Let me know if i missed anything here.
     
  2. jcsd
  3. Aug 22, 2010 #2
    It's not constant coefficient but you can transform it to a constant coefficient by your change of independent variable [itex]x=e^t[/itex]. This is also an Euler-type equation, the homogeneous part is more easily solved by letting [itex]y=x^k[/itex], substituting, then figuring out what k is. That gives you the [itex]y_c=a/x+bx^{-8}[/itex] part. However I assume you need to do the derivative substitutions where:

    [tex]\frac{dy}{dx}=e^{-t}\frac{dy}{dt}[/tex]

    [tex]\frac{d^2y}{dx^2}=e^{-2t}\left(\frac{d^2 y}{dt^2}-\frac{dy}{dt}\right)[/tex]

    You can obtain those right? Then just substitute into the equation to obtain the constant-coefficient equation in the variable t, then let t=ln(x) to get it back into x


    Then you need to solve the nonhomogeneous part which I guess reduction of order is the easiest approach.
     
  4. Aug 22, 2010 #3

    LCKurtz

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    It would probably be easier to solve the NH equation while it is in terms of t using undetermined coefficients and transfer the whole thing back to x for the general solution.
     
  5. Aug 22, 2010 #4
    Thanks jackmell and LCKurtz

    I changed the independent variable to x=et. Yes, jackmell either solution works for the homogenous part.

    However, I am stuck with x2/30 which I believe is the particular solution. i.e. I had set G(x) = x2 where the partcular solution is of the form Ax2+Bx+C.

    jackmell, LCKurtz -> please could you elaborate your answers further.

    Thanks for taking the time to respond.
     
  6. Aug 22, 2010 #5

    LCKurtz

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    After you made the x = et substitution, presumably you got the new DE:

    [tex]\frac{d^2y}{dt^2}+ 9\frac{dy}{dt} + 8y = e^{2t}[/tex]

    which has complementary solution

    [tex]y_c=Ae^{-8t}+Be^{-t}[/tex]

    Use undetermined coefficients to find a particular solution of this NH equation in t. Given you have e2t on the right you would look for a particular solution of the form

    [tex]y_p = Ce^{2t}[/tex]

    Figure out C and you have the general solution y = yc + yp, all expressed in terms of t. Then substitute back for x to get the general solution and get your missing x2/30 term.
     
  7. Aug 24, 2010 #6
    Hi LCKurtz

    got it. thanks for the help and your time.
     
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