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Math Help - Second-Order Linear Non Homogeneous ODE with Constant Coefficients

  1. #1
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    Sep 2009
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    [SOLVED] Second-Order Linear Non Homogeneous ODE with Constant Coefficients

    Here it is:
    <br />
\frac{d^2y}{dx^2}+6\frac{dy}{dx}+25y=-292sin(4x)<br />

    So the auxiliary/characteristic equation will look like:
    <br />
r^2+6r+25=0 <br />
    Solving that gives:
    <br />
 r^2+6r+25=0<br />
    <br />
r = 3\pm4i<br />

    And so we'll get the following general homogeneous solution:
    <br />
  GS(H) = e^{3x}(Acos(4x)+Bsin(4x))<br />

    So now I try to find the particular non homogeneous solution, trying:
    <br />
  y_P = \alpha xcos(4x)+\beta xsin(4x)<br />
    Finding the first and second derivative of y particular:
    <br />
   y_P' = \alpha cos(4x) - 4\alpha xsin(4x) + \beta sin(4x) + 4\beta xcos(4x)<br />
    <br />
y_P'' = -8\alpha sin(4x) + 8\beta cos(4x) - 16\alpha xcos(4x) - 16\beta xsin(4x)<br />

    Then, when I try to compile all of this, to then equate coefficients and solve for alpha and beta, I realise that the x terms do not cancel out (as they should) and I cannot go any further (I don't think).

    It will eventually look like this:
    <br />
-8\alpha sin(4x) + 8\beta cos(4x) - 16\alpha xcos(4x) - 16\beta xsin(4x)<br />
    <br />
+ 6\alpha cos(4x) - 24\alpha xsin(4x) + 6\beta sin(4x) + 24\beta xcos(4x)<br />
    <br />
 + 25\alpha xcos(4x) + 25\beta xsin(4x) = -292sin(4x)<br />


    Have I made an error somewhere?

    Any help is appreciated.


    NB: It may/may not help, but in the previous part of the question we've already proved that when:
    <br />
x=0, \frac{dy}{dx}=-2<br />
    Last edited by yianni; October 22nd 2009 at 11:14 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by yianni View Post
    Here it is:
    <br />
\frac{d^2y}{dx^2}+6\frac{dy}{dx}+25y=-292sin(4x)<br />

    So the auxiliary/characteristic equation will look like:
    <br />
r^2+6r+25=0 <br />
    Solving that gives:
    <br />
r^2+6r+25=0<br />
    <br />
r = 3\pm4i<br />

    And so we'll get the following general homogeneous solution:
    <br />
GS(H) = e^{3x}(Acos(4x)+Bsin(4x))<br />

    So now I try to find the particular non homogeneous solution, trying:
    <br />
y_P = \alpha xcos(4x)+\beta xsin(4x)<br />
    Why is the trial particular solution not:

    y_P = \alpha \cos(4x)+\beta \sin(4x)

    CB
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  3. #3
    MHF Contributor

    Joined
    Apr 2005
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    1327
    Quote Originally Posted by yianni View Post
    Here it is:
    <br />
\frac{d^2y}{dx^2}+6\frac{dy}{dx}+25y=-292sin(4x)<br />

    So the auxiliary/characteristic equation will look like:
    <br />
r^2+6r+25=0 <br />
    Solving that gives:
    <br />
 r^2+6r+25=0<br />
    <br />
r = 3\pm4i<br />

    And so we'll get the following general homogeneous solution:
    <br />
  GS(H) = e^{3x}(Acos(4x)+Bsin(4x))<br />

    So now I try to find the particular non homogeneous solution, trying:
    <br />
  y_P = \alpha xcos(4x)+\beta xsin(4x)<br />
    I suspect you put that "x" multiplying the sine and cosine in because you saw "(A cos(4x)+ Bsin(4x))" in the homogeneous solution. However, you really have e^{3x}cos(4x) and e^{3x}sin(4x) which are NOT equivalent to "cos(4x)" and "sin(4x)". More simply, the solutions to you characteristic equation are 3+ 4i and 3- 4i while just sin(4x) and cos(4x) correspond to 4i and -4i- quite different solutions.

    Finding the first and second derivative of y particular:
    <br />
   y_P' = \alpha cos(4x) - 4\alpha xsin(4x) + \beta sin(4x) + 4\beta xcos(4x)<br />
    <br />
y_P'' = -8\alpha sin(4x) + 8\beta cos(4x) - 16\alpha xcos(4x) - 16\beta xsin(4x)<br />

    Then, when I try to compile all of this, to then equate coefficients and solve for alpha and beta, I realise that the x terms do not cancel out (as they should) and I cannot go any further (I don't think).

    It will eventually look like this:
    <br />
-8\alpha sin(4x) + 8\beta cos(4x) - 16\alpha xcos(4x) - 16\beta xsin(4x)<br />
    <br />
+ 6\alpha cos(4x) - 24\alpha xsin(4x) + 6\beta sin(4x) + 24\beta xcos(4x)<br />
    <br />
 + 25\alpha xcos(4x) + 25\beta xsin(4x) = -292sin(4x)<br />


    Have I made an error somewhere?

    Any help is appreciated.


    NB: It may/may not help, but in the previous part of the question we've already proved that when:
    <br />
x=0, \frac{dy}{dx}=-2<br />
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  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    3
    Ah, I see my mistake.

    Thanks guys; solved.
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