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Expressing a DE as two coupled first order DEs Watch

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    How do you convert a second or higher order differential equation into a system of coupled first order differential equations?

    e.g.



    My lecturer said that a standard way of doing this for a second order differential equation is let
    and but he didn't explain a way of doing this.

    For the above DE, I tried using
    and and then substituting these, rearranging all terms to one side and then setting sets of terms equal to zero to establish the answers:

    and .

    Are these correct? If not, what is a better way of approaching it?
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    (Original post by OL1V3R)
    How do you convert a second or higher order differential equation into a system of coupled first order differential equations?

    e.g.



    My lecturer said that a standard way of doing this for a second order differential equation is let
    and but he didn't explain a way of doing this.

    For the above DE, I tried using
    and and then substituting these, rearranging all terms to one side and then setting sets of terms equal to zero to establish the answers:

    and .

    Are these correct? If not, what is a better way of approaching it?
    The method you are trying to use will not work in this case, because it is not an ODE in dy/dt i.e. you have a y term.

    I'm not too sure how to solve this tbh.
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    In general, if you have a n-th order linear ODE you need n coupled first-order ODEs to express it.
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    I found an answer to this, I sent my lecturer an email and it was a lot easier than I thought!

    In this case, if you let f = dy/dt, g=df/dt, and h = dg/dt, then h = t + 2y + 3tf - tg. That's how I was told to do it. For our exam though, we don't need to solve it, just express the original DE as coupled first order DEs.
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    If you have an nth-order differential equation L(y,y',y'',\dots, y^{(n-1)}, y^{(n)})=f(x) (where L is linear), then you can let g_1=y',\ g_2=g_1' (= y''),\ \dots,\ g_{n-1} = g'_{n-2} (=y^{(n-1)}), and then L(y,g_1,g_2, \dots, g_{n-1}, g'_{n-1}) = f(x). These substitution formulae and the substitution into the original equation then constitute the n linear differential equations in question, i.e.:

    \begin{cases} y'-g_1=0 \\ g'_1-g_2=0 \\ \vdots \\ g'_{n-2} - g_{n-1}=0 \\ L(y,g_1, g_2, \dots, g_{n-1}, g'_{n-1}) = f(x) \end{cases}
 
 
 
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