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Math Help - Two functions that simultaneously satisfy the system of equations

  1. #1
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    Two functions that simultaneously satisfy the system of equations

    y = Y(x) and z = Z(x) simultaneously satisfy the system of equation:

    y' = z, z' = x^3 (y = z)

    The problem gives you initial conditions and initial guesses (for successive approximation). But my question is how can I transform this into y wrt x and z wrt x without going to second order. Because the formula that I know for doing approximation involves only two variable and only to the first order.
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  2. #2
    Super Member Rebesques's Avatar
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    how can I transform this into y wrt x and z wrt x without going to second order
    You can't.
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  3. #3
    A Plied Mathematician
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    Why not just integrate the z equation first, obtaining

    z(x)=\dfrac{x^{4}}{4}+C, and then integrate the y equation, getting

    y(x)=\dfrac{x^{5}}{20}+Cx+D?

    Kinda simplistic, but it gets the job done, doesn't it? That is, if you're trying to solve the DE exactly, here's the exact solution. If you're doing successive approximations, that's something else.
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  4. #4
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    Quote Originally Posted by hashshashin715 View Post
    y = Y(x) and z = Z(x) simultaneously satisfy the system of equation:

    y' = z, z' = x^3 (y = z)

    The problem gives you initial conditions and initial guesses (for successive approximation). But my question is how can I transform this into y wrt x and z wrt x without going to second order. Because the formula that I know for doing approximation involves only two variable and only to the first order.
    As Ackbeet said, it is easy to solve these equations, just by integrating. But what exactly are you trying to do? Apparently use some kind of approximation method? What method?
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  5. #5
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    son of a bum, i put a typo in my original post. It should be:

    y' = z
    z' = (x^3)(y + z)

    but my class starts in 2 hours and i wont be able to check MHF, so thanks for trying.
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  6. #6
    A Plied Mathematician
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    You might try matrix methods. The system becomes

    \begin{bmatrix}y'\\z'\end{bmatrix}=\begin{bmatrix}<br />
0 &1\\<br />
x^{3} &x^{3}<br />
\end{bmatrix}\begin{bmatrix}y\\z\end{bmatrix}.

    The trivial solution obviously works here. Do you have any initial conditions? You might have a hard time coming up with a fundamental matrix.

    Any particular reason you have to avoid the second-order method?
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