Reducing the degree of an ODE by substitution question

Question

I have a fairly simple question but I'm new to ODEs so it has been bugging me and I haven't been able to find a solution nor check my own.

So, in my workbook we have instructions on solving different types of higher-level ODEs.

If an ODE is in the following form: $$F(y', y'', \dots, y^{(n)})$$ we can use the transformation $y' = p$, where $y'' = p \frac{dp}{dy} = p'p$, after which we substitute those back into the equation. This is the case where the ODE does not contain the independent variable X.

However, I have a problem when I want to figure out $y'''$ and so on. I don't know how to differentiate $y''$ with respect to the corresponding variable y. I would appreciate any help!

Answer 1

When you substitute $y' = p$, you get $y'' = (y')' = p'$, not $y'' = p' p$. For higher orders, you can then continue to introduce variables representing the derivative of that order, i.e., $y' = p_1$, $y'' = p_2$, and so on. To reduce an ODE of order $n$ to one of order $1$, you will need to introduce $n-1$ additional variables. This increases the dimensionality of your problem, but numerical methods can handle that pretty well.

Answer 2

$$y'=\dfrac {dy}{dx}=p$$ Just differentiate and apply the chain rule: $$y''=\dfrac {dy'}{dx}=\dfrac {dp}{dx}=\dfrac {dp}{dy}\dfrac {dy}{dx}=p'p$$ Apply the same rules for $y'''$:

$$ \begin{align} y'''&=\dfrac {dy''}{dx} =\dfrac {dy''}{dy}\dfrac {dy}{dx} \\ y'''&=p\dfrac {dy''}{dy} \\ y'''&=p\dfrac {d}{dy}(pp') \\ y'''&=p \left(p\dfrac {dp'}{dy}+p'\dfrac {dp}{dy}\right) \\ y'''&=p(pp''+p'^2) \\ y'''&=p^2p''+pp'^2 \\ \end{align} $$ Where $p=\dfrac {dy}{dx}$ and $p'=\dfrac {dp}{dy}$ and $p''=\dfrac {d^2p}{dy^2}$