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# Expressing a DE as two coupled first order DEs Watch

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

e.g.

$\frac{\textup{d}^{3}y}{\textup{d}t^{3}}+t\frac{\textup{d}^{2}y}{\textup{d}t^{2}}-3t\frac{\textup{d}y}{\textup{d}t}-2y=t$

My lecturer said that a standard way of doing this for a second order differential equation is let
$g=\frac{\textup{d}y}{\textup{d}t}$ and $\frac{\textup{d}g}{\textup{d}t}=\frac{\textup{d}^{2}y}{\textup{d}t^{2}}$ but he didn't explain a way of doing this.

For the above DE, I tried using
$g=\frac{\textup{d}^{2}y}{\textup{d}t^{2}}$ and $\frac{\textup{d}g}{\textup{d}t}=\frac{\textup{d}^{3}y}{\textup{d}t^{3}}$ and then substituting these, rearranging all terms to one side and then setting sets of terms equal to zero to establish the answers:

$\frac{\textup{d}g}{\textup{d}t}=(1-g)t$ and $\frac{\textup{d}y}{\textup{d}t}=-\frac{2y}{3t}$.

Are these correct? If not, what is a better way of approaching it?
2. (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.

$\frac{\textup{d}^{3}y}{\textup{d}t^{3}}+t\frac{\textup{d}^{2}y}{\textup{d}t^{2}}-3t\frac{\textup{d}y}{\textup{d}t}-2y=t$

My lecturer said that a standard way of doing this for a second order differential equation is let
$g=\frac{\textup{d}y}{\textup{d}t}$ and $\frac{\textup{d}g}{\textup{d}t}=\frac{\textup{d}^{2}y}{\textup{d}t^{2}}$ but he didn't explain a way of doing this.

For the above DE, I tried using
$g=\frac{\textup{d}^{2}y}{\textup{d}t^{2}}$ and $\frac{\textup{d}g}{\textup{d}t}=\frac{\textup{d}^{3}y}{\textup{d}t^{3}}$ and then substituting these, rearranging all terms to one side and then setting sets of terms equal to zero to establish the answers:

$\frac{\textup{d}g}{\textup{d}t}=(1-g)t$ and $\frac{\textup{d}y}{\textup{d}t}=-\frac{2y}{3t}$.

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.
3. In general, if you have a n-th order linear ODE you need n coupled first-order ODEs to express it.
4. 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.
5. If you have an nth-order differential equation (where L is linear), then you can let , and then . These substitution formulae and the substitution into the original equation then constitute the n linear differential equations in question, i.e.:

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