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Math Help - System of Linear Differential Equations with Constant Coefficients

  1. #1
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    System of Linear Differential Equations with Constant Coefficients

    Hello Everyone,

    Now I am familiar, roughly, in solving systems of DEs with constant coefficients, granted that these equations are homogenous, but how can we solve such a system of equations:

    y_1'=ay_1+by_2<br />
y_2'=cy_1+dy_2+f(x)

    Thanks a lot!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    You have to find a particular solution of the complete system and add all the solutions of the homogeneous. There is a well known theorem about it. Have you studied the corresponding theory?
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    I am afraid not, I konly know how to solve linear ODEs (not necessarily homogenous) and homogenous systems. I tried searching for an answer but didn't know what to search for.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by rebghb View Post
    I am afraid not, I konly know how to solve linear ODEs (not necessarily homogenous) and homogenous systems. I tried searching for an answer but didn't know what to search for.

    I suppose you know how to compute the exponential of a matrix. A particular solution for the system

    Y'=AY+b(x)

    is

    Y(x)=\displaystyle\int_0^xe^{tA}b(t)dt
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  5. #5
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    Quote Originally Posted by rebghb View Post
    Hello Everyone,

    Now I am familiar, roughly, in solving systems of DEs with constant coefficients, granted that these equations are homogenous, but how can we solve such a system of equations:

    y_1'=ay_1+by_2<br />
y_2'=cy_1+dy_2+f(x)

    Thanks a lot!
    Well, no, the second equaton is NOT homogeneous.

    Here is a much more elementary method:

    Differentiate both sides of the first equation to get
    y_1''= ay_1'+ by_2'
    From the second equation, by_2'= b(cy_1+ dy_2+ f(x))= bcy_1+ bdy_2+ bf(x) so we can write that as
    y_1''= ay_1'+ bcy_1+ bdy_2+ bf(x)
    and now, from the first equation again, by_2= y1'- ay_1 so we have
    y_1''= ay_1'+ bcy_1+ d(y_1'- ay_1)+ bf(x)
    y_1''= (a+ d)y_1'+ (a-ad)y_1+ bf(x)
    a single, non-homogeneous second order linear equation with constant coefficients. Once you have solved for y_1 you can use by_2= y_1'- ay_1 to find = y_2
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  6. #6
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    Thank you, I thought of that, honestly, but what I was looking for was something to solve a bigger problem, say 10 equations!
    As I have posted earlier, we need to find the eigenvalues for the matrix, a step towards solving the system, shall the system be homogenous; but what if it isn't? Is there a particular procedure? If yes I'd like to know it's name, I'll try to search for it and learn it on my own...
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    A Plied Mathematician
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    Quote Originally Posted by FernandoRevilla View Post
    I suppose you know how to compute the exponential of a matrix. A particular solution for the system

    Y'=AY+b(x)

    is

    Y(x)=\displaystyle\int_0^xe^{tA}b(t)dt
    Did you mean

    \displaystyle Y(x)=e^{tA}\int_{0}^{x}e^{-tA}b(x)\,dx?

    [EDIT]: See Fernando's next post for a correction in the variables.
    Last edited by Ackbeet; February 28th 2011 at 04:52 AM. Reason: Fernando correction.
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Did you mean

    \displaystyle Y(x)=e^{tA}\int_{0}^{x}e^{-tA}b(x)\,dx?

    No, I meant:


    Y(x)=e^{xA}\displaystyle\int_0^xe^{-tA}b(t)\;dt

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  9. #9
    A Plied Mathematician
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    Yeah, that. Right.
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