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Thread: solving a system by substitution 3b

  1. #1
    MHF Contributor
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    solving a system by substitution 3b

    solve this system by sustituting$\displaystyle t=e^u$ and building a function $\displaystyle y(u)=x(e^u)$
    $\displaystyle
    t\frac{dx}{dt}=(\begin{array}{cc}
    2 & 2\\
    1 & 3\end{array})x
    $
    $\displaystyle
    x(1)=(\begin{array}{c}
    2\\
    3\end{array})
    $

    how to do it?
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  2. #2
    Super Member PaulRS's Avatar
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    Apply the chain rule to get: $\displaystyle \displaystyle \frac{dy}{du} = e^u \cdot \frac{dx}{dt} = t \cdot \frac{dx}{dt}$ ( $\displaystyle \frac{dx}{dt}$ evaluated at $\displaystyle e^u = t$ )

    Then you have $\displaystyle \dot{y} = \begin{pmatrix}
    2 & 2\\
    1 & 3
    \end{pmatrix} \cdot y$ now you could solve it, say, by computing the matrix exponential.

    For the initial values think what knowing $\displaystyle x(1)$ would mean in terms of $\displaystyle y$, at the end, when you have $\displaystyle y$ you can undo the change of variables to get the solution for $\displaystyle t>0$.
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  3. #3
    MHF Contributor
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    you sat that y=x(u)t(u)=x(u)e^y
    so the derivative is must have sum of two member
    why you have one?
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