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Homework Help: Solving for an unknown in differential equation solution

  1. Aug 6, 2009 #1
    1. The problem statement, all variables and given/known data

    (1): [tex]\frac{d^{2}y}{dx^{2}} + 4\frac{dy}{dx} + 20y = 10e^{-2x}\cos (rx)[/tex], [tex]y(0) = 1[/tex], [tex]y'(0) = 0[/tex]

    Show that there is precisely one value for r for which [tex] e^{2x}y(x)[/tex] can become arbitrarily large where [tex]y(x)[/tex] is a solution of (1)

    3. The attempt at a solution

    I used Maple to solve the DE and found a particular solution [tex]y(x)[/tex].

    [tex] e^{2x}y(x) = -\frac{10(-\cos 4x + \cos rx)}{-16+r^2}[/tex]

    When r tends to 4 or -4 then [tex]e^{2x}y(x)[/tex] can become arbitrarily large, but this doesn't seem to me like it satisfies the question - it's not "precisely one value" and it's not an exact value of r. I graphed [tex] e^{2x}y(x)[/tex] and can't see anywhere else it can become arbitrarily large. I feel like I'm missing something, can anyone point me in the right direction? Cheers.
  2. jcsd
  3. Aug 6, 2009 #2


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    Hi Ewan_C! :wink:

    Sorry, but this is what happens when you take short-cuts …

    there is a solution for r = ±4, but it doesn't involve cos4x.

    Start again, put y = ze-2x, get the differential equation for z, put r = ±4, and look for another particular solution. :smile:
  4. Aug 6, 2009 #3
    Hi Tim, thanks for the response.

    So you mean that I should substitute y(x) = z(x)*e-2x into the original DE and solve it for z? Now I get it, that's what they meant from the start - I took "where y(x) is a solution of (1)" to mean a particular solution rather than the general solution.

    We haven't yet been taught how to solve DEs with non-constant coefficients, so I used Maple again to solve the DE. I used the following commands:

    DE := diff(exp(-2*x)*y(x),x,x)+4*diff(exp(-2*x)*y(x),x)+20*exp(-2*x)*y(x) =10*exp(-2*x)*cos(r*x);

    DE_hom := diff(exp(-2*x)*y(x),x,x)+4*diff(exp(-2*x)*y(x),x)+20*exp(-2*x)*y(x) = 0;

    dsolve({DE,y(0)=1,D(y)(0)=0},y(x)) - dsolve({DE_hom,y(0)=1,D(y)(0)=0},y(x));

    I didn't substitute y(x)=z(x)e-2x as it made Maple get angry with me - my Maple skills aren't too brilliant! I think what I did is equivalent though?
    Am I correct in thinking that if you take the general solution of the DE and subtract the general solution of the corresponding homogeneous DE, the result is a particular solution of the DE?

    This still gave me a particular solution with a cos(4x) term. I don't really know what's going on with this DE as I can't solve it by hand. Maybe the best next step would be to spend some time on the internet learning how solve DEs with non-constant coefficients?
  5. Aug 6, 2009 #4


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    Sorry, Ewan_C, I've no idea what Maple is or how it works. :redface:

    Surely you can work out what dy/dx and d2y/dx2 are in terms of dz/dx and d2z/dx2 without a calculator? :smile:
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