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Math Help - diffrential equations

  1. #1
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    diffrential equations

    I have this problem to solve the diffrential equation, can someone give me a hint where I should start
    \left\{\begin{array}{cc}x'+5x-2y=e^t\\y'-x+6y=e^{-2t}\end{array}\right.
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  2. #2
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    This is easily solved by using methods from linear algebra. The system can be written as:

    \left(\begin{array}{c}x_1\\x_2\end{array}\right)'=  \left(\begin{array}{cc}-5&2\\1&-6\end{array}\right)\left(\begin{array}{c}x_1\\x_2\  end{array}\right)+\left(\begin{array}{c}e^t\\e^{-2t}\end{array}\right)

    The eigenvalues that belong to the transformation matrix are -4 and -7, so you end up with the eigenvectors:

    T=\left(\begin{array}{cc}2&1\\1&-1\end{array}\right)

    With inverse:

    T^{-1}=\frac{1}{3}\left(\begin{array}{cc}1&1\\1&-2\end{array}\right)

    The new transformation matrix becomes: A_f=\left(\begin{array}{cc}-4&0\\0&-7\end{array}\right)

    And the system is transformed to:

    \left(\begin{array}{c}y_1\\y_2\end{array}\right)' = A_fY + T^{-1}\left(\begin{array}{c}e^t\\e^{-2t}\end{array}\right)=\left(\begin{array}{cc}-4&0\\0&-7\end{array}\right)\left(\begin{array}{c}y_1\\y_2\  end{array}\right)+\frac{1}{3}\left(\begin{array}{c  c}1&1\\1&-2\end{array}\right)\left(\begin{array}{c}e^t\\e^{-2t}\end{array}\right)

    Then just solve the two systems independently and transform the solution to the old coordinates by using: X=TY
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  3. #3
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    I have only done Linear Algebra 103!

    I don't understand this, sorry I wasn't clear enough about my math skills
    This is in a chapter called "simultaneous differential equations".
    So I can not return the solutions in linear algebric form, my teacher would defenitly see that I got some advanced help
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  4. #4
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    Quote Originally Posted by hlolli View Post
    I have only done Linear Algebra 103!

    I don't understand this, sorry I wasn't clear enough about my math skills
    This is in a chapter called "simultaneous differential equations".
    So I can not return the solutions in linear algebric form, my teacher would defenitly see that I got some advanced help
    In that case, from the second equation:

    x = y' + 6y - e^{-2t} .... (A)

    \Rightarrow x' = y'' + 6y' + 2 e^{-2t} .... (B)

    Substitute (A) and (B) into the first equation and re-arrange:

    y'' + 11 y' + 28 y = 3 e^{-2t} + e^t .... (C)

    Solve (C) for y = y(t). Substitute the solution into (A) to get x = x(t).
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    In that case, from the second equation:

    x = y' + 6y - e^{-2t} .... (A)

    \Rightarrow x' = y'' + 6y' + 2 e^{-2t} .... (B)

    Substitute (A) and (B) into the first equation and re-arrange:

    y'' + 11 y' + 28 y = 3 e^{-2t} + e^t .... (C)

    Solve (C) for y = y(t). Substitute the solution into (A) to get x = x(t).
    Thank you, after a few hours I figured it out and I got the same. Now I know I did right!
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