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Math Help - Non-homogeneous system of ODEs (need particular solution)

  1. #1
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    Non-homogeneous system of ODEs (need particular solution)

    Hi,
    I need to find the particular solution to the following system of equations using the method of undetermined coefficients.

    dx/dt = y + e^t

    dy/dt = -2x + 3y + 1

    The problem is the forcing terms of the two diff equations are different, that is one is polynomial and the other an exponential. I have no idea what you "guess" the particular solution set to be in this situation.

    Any help will be GREATLY appreciated!

    Thank you!
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  2. #2
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    You can decouple them right?

    x'=y+e^2

    y'=-2x+3y+1\Rightarrow y''=-2x'+3y'+1

    or:

    y''-3y'+2y=1-2e^t

    Solve that one, plug it into the first one, then solve that one. Looks to me anyway.
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  3. #3
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    yes shawsend, that can be done.. but I really doubt we're allowed to change the diff equations to higher orders.. that's how the rest of the Qs were done.

    For example if you had a system like
    dx/dt = x + 3y + 2e^(4t)
    dy/dt = 2x + 2y - e^(4t)
    the particular solution would be of the form c*t*w*e^(4t) + u*e^(4t) where w is an eigenvector of the coefficient matrix of the system and 4 is an eigenvalue. The "coefficients" we have to find are c and the elements of vector u.. say u = (a,b)^T.. so a, b and c
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  4. #4
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    Yea, Ok. Need to look at it this way then:

    \binom{x}{y}^{'}=\left(\begin{array}{cc}\phantom{-}0&1\\-2&3\end{array}\right)\binom{x}{y}+\binom{e^t}{1}

    I know, that doesn't help you but it looks nice I think. First step as you know is to find the general solution for homogeneous system. Got that?

    This problem is in Rainville and Bedient, p. 270. It's a good book on differential equations I think. That and Blanchard, Hall and Devaney if you're serious about DEs. Couple more.

    What's the next step in matrix form?
    Last edited by shawsend; October 1st 2008 at 08:16 AM.
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  5. #5
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    yeah thanks, that's what i was talking about.

    I got the homogeneous solution.

    But the problem is I've only ever done systems when the forcing terms are the same term.. like 2e^(2t) and -5e^(2t). Then I can guess the particular solution and find the coefficients.

    But in this case the two forcing terms are of different functional form. i.e. one polynomial and one exponential... what would the particular solution be then?

    Also, does Rainville and Bedient give a full solution to this problem?

    thanks
    Last edited by hashi; October 1st 2008 at 08:35 PM.
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  6. #6
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    So the general solution to the homogeneous system is:

    \binom{x}{y}_{\hspace{-3pt}c}=a_1\binom{1}{1}e^{t}+a_2\binom{1}{2}e^{2t}

    Now we do what the alien told professor Barnard in ``The day the earth stood still'': use variation of parameters. We then seek a solution of the form:

    \binom{x}{y}_{\hspace{-3pt}p}=a_1(t)\binom{1}{1}e^{t}+a_2(t)\binom{1}{2}e  ^{2t}

    Substitute this particular solution into the original DE:

    a'_1(t)\binom{1}{1}e^t+a'_2(t)\binom{1}{2}e^{2t}=\  binom{e^t}{1}

    You can now use Cramer's rule to find a'_1(t) and a'_2(t).

    Yes, the problem is solved in Rainville and Bedient. Blanchard, Devaney and Hall go into great detail of systems offering insight into the underlying dynamics; very nice perspective of the world that few see in my opinion.
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  7. #7
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    can it also be done using the method of undetermined coefficients? Coz that's what the question asks us to do "use the method of undetermined coefficients"

    Thanks
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  8. #8
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    Quote Originally Posted by hashi View Post
    can it also be done using the method of undetermined coefficients?
    Thanks
    uh . . . got Kreyszig? That's another one I think is good to have. You know, "Advanced Engineering Mathematics". Anyway, he goes through an example using undetermined coefficients, p. 184. Would take time for me also to review it. Hopefully you can find the book in the library and get two birds with one stone (it's a good book to know about, and solve the problem).
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  9. #9
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    Here's my solution via undetermined coefficients. Can't say I'm tops with this but this is what I got: The homogeneous analog of equation:
    <br />
\textbf{X}'=\textbf{A}\textbf{X}+\textbf{G}=\left(  \begin{array}{cc}\phantom{-}0 & 1 \\ -2&3\end{array}\right)\binom{x}{y}+\binom{e^t}{1}
    has the solution:
    \binom{x}{y}_c=a_1\binom{1}{1}e^t+a_2\binom{1}{2}e  ^{2t}

    Note that \lambda=1 is an eigenvalue of the coefficient matrix and the non-homogeneous part has an e^t factor. Thus like repeated roots in an ordinary ODE, we must consider te^t as a component of the particular solution as well as e^t and a constant vector. Thus I'll solve for the column vectors of:

    y_p=\binom{x}{y}_p=\textbf{U}+\textbf{V}e^t+\textb  f{W}te^t=\binom{u_1}{u_2}+\binom{v_1}{v_2}e^t+\bin  om{w_1}{w_2}te^t. Substituting this into the DE:

    \textbf{V}e^t+\textbf{W}(te^t+e^t)=\textbf{A}\text  bf{U}+\textbf{A}\textbf{V}e^t+\textbf{A}\textbf{W}  te^t+\binom{e^t}{1}

    Equating coefficients for the constant terms, e^t, and te^t terms I get the following sets of equations:

    \begin{aligned}u_2&=0 &  w_1&=w2 & v_1+w_1&=v_2+1 \\<br />
               2u_1+3u_2&=-1 & w_2&=-2w_1+3w_2 & v_2+w_2&=-2v_1+3v_2<br />
               \end{aligned}<br />

    From these I can write a particular solution as:

    y_p=\binom{1/2}{0}+\binom{0}{1}e^t+\binom{2}{2}te^t
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  10. #10
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    ok that helped heaps and heaps!!

    Thank you very very much!!!!
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  11. #11
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    Hello..I have to solve this question..actually I don't know how I can do this kind of exercise with three parameters.



    Reduce the system of linear higher order equations

    sint(d^2x/dt^2)-3(dy/dt)+2(e^t)=tanht

    2(d^2y/dt^2)+4(t^1/2)(dx/dt)-9x+4y=cosht

    to a linear system of first order equations. Make sure to give the matrix A(t) and the
    inhomogeneous term H(t).



    Any help it will be appreciated..

    Thank you
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