# Reducing the degree of an ODE by substitution 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!

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.
$$y'=\dfrac {dy}{dx}=p$$ 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 $$p''=\dfrac {d^2p}{dy^2}$$