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Homework Help: General solution ODE

  1. Aug 5, 2010 #1
    Im having trouble with this question. can anyone explain please?

    1. The problem statement, all variables and given/known data
    y'' + 6y' + 9y = x*exp(-3x)3x

    2. Relevant equations

    Find the general solution.
  2. jcsd
  3. Aug 5, 2010 #2
    Well, the general solution is the sum of the homogeneous solution and a particular solution.

    The homogeneous solution (gotten from putting the right hand side = 0) is obtained using the characteristic equation.

    To find a particular solution, one can make a clever guess how the solution must look like. Since the right hand side has a part [tex]e^{-3x}[/tex] one can safely assume that also [tex]y[/tex] must have this part. To deal with the [tex]x[/tex] one can guess that the solution must contain a polynomial of some degree. Thus, a clever guess would be [tex]y=p(x)e^{-3x}[/tex]. All that is left is to see which [tex]p(x)[/tex] that will satisfy the solution.
  4. Aug 5, 2010 #3
    thanks for the reply.

    im working through this question. im not to sure im doing it right though.

    K^2+6K+9 = 0
    therefore K= -3
    and y= A*exp(-3x) + Bx*exp(-3x)
    y'=-3A*exp(-3x) - 3Bx*exp(-3x) + B*exp(-3x)

    y(p)= Ax*exp(-3x)
    y'(p)= A*exp(-3x) - 3Ax*exp(-3x)

    am i on the right track?
  5. Aug 5, 2010 #4
    The homogeneous solution looks correct, although you do not have to differentiate it. You merely have to add the solutions together in the end, [tex]y=y_H+y_{P}[/tex], so you are done with [tex]y_H[/tex].

    In the particular solution you have assumed that the polynomial is [tex]p(x)=Ax[/tex], which is not the general case. In fact, a polynomial of larger order is needed. I think it is easier if you treat [tex]p(x)[/tex] as a general polynomial, and then you will get an expression for the polynomial in the end, when you have put in [tex]y=p(x)e^{-3x}[/tex] in the differential equation.
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