Background
I'm currently reviewing for the final exam of my undergraduate Ordinary Differential Equations course, and a question on the last exam that stumped me was the following:
Find the general solution for the ODE $ y'' = -2t(y')^2 $.
Although I unfortunately couldn't solve it during my mid-exam panic, I eventually went home and solved it the following way:
Let $ v = y' \implies v' = y'' $. Apply this substitution to the above equation.
Using the separation of variables method, turn $ v' = -2tv^2 $ into $ v^{-2}v' = -2t $.
$$ \int{v^{-2}dv} = \int{-2tdt} $$ $$ -v^{-1} = -t^2 + c, c \in \Re $$ $$ v^{-1} = t^2 - c $$ $$ v = \frac{1}{t^2 - c} $$ $$ y' = \frac{1}{t^2 - c} $$ $$ \int{1dy} = \int{\frac{1}{t^2 - c}dt} $$ $$ y = \int{\frac{1}{t^2 - c}dt} $$
Finally, I'd evaluate the integral for a negative, zero, and positive $ c $ to get the final general solution to the ODE, but that part is beside the point of the question.
Problem
My request is for someone to provide several similar examples of higher order, nonlinear ODEs that can be solved through a similar method of substitution as above, as well as some ODEs that require substitutions to be made in order prior to solving them.
It's hard to find similar examples online, though few are present in undergraduate textbooks, so I'm hoping people reading this would find the upcoming solutions useful.
Thank you in advance!
The method of solution you have used works whenever the unknown function $y$ does not appear explicitly in the equation, that is, in equations of the form $$ y''=f(t,y'). $$ The change $y'=v$ reduces it to the first order ODE $$ v'=f(t,v). $$
Another situation in which this can be done is when the independent variable $t$ does not appear explicitly in the equation, that is, equations of the form $$ y''=f(y,y'). $$ Let $y'=p$ and consider $p$ as a function of $y$. Then $$ y''=\frac{dy'}{dt}=\frac{dp}{dt}=\frac{dp}{dy}\,\frac{dy}{dt}=p\,\frac{dp}{dy}. $$ The original equation becomes $$ p\,\frac{dp}{dy}=f(y,p), $$ a first order equation. Once you find $p=p(y)$, you still have to find $y$ as a function of $t$. This is done solving the equation $$ \frac{dy}{dt}=p(y), $$ whose solution is $$ \int\frac{dy}{p(y)}=t+C. $$