For constants $a,b$ there are many ways to find the solutions to
$$y^{\prime\prime} + (a+b)y^\prime + aby = \phi(x). $$
Perhaps the most popular is to first solve the homogeneous case when $\phi(x) = 0$ and then find a particular solution using the 'guess and check' method.
There seems to be an easier way by solving this as two first order ODEs, and my questions are:
- is this way easier (in your opinion)?
- if so, why do we not teach people to do this the easier way? And,
- does the method described below have a name?
I claim the easier way is to make the substitution $u = y^\prime + ay$. Then the original 2nd order ODE can be written as a first order linear ODE $$ u^\prime + bu = \phi(x). $$ Once we solve this equation for $u$, we then return to our substitution $$ y^\prime + ay = u.$$ The solutions to this first order linear ODE are the solutions to the original second order ODE.
Here is a YouTube clip of me explaining this method in the case when $a=b$.
This is my preferred method. In general, let $$ \partial _{x}^{2}y-c\partial _{x}y-dy=\varphi (x) $$ Set $$ y_{1}=y,\;y_{2}=\partial _{x}y $$ Then \begin{eqnarray*} \partial _{x}y_{1} &=&y_{2} \\ \partial _{x}y_{2} &=&dy_{1}+cy_{2}+\varphi (x) \end{eqnarray*} or $$ \partial _{x}\left( \begin{array}{c} y_{1} \\ y_{2} \end{array} \right) =\left( \begin{array}{cc} 0 & 1 \\ d & c \end{array} \right) \left( \begin{array}{c} y_{1} \\ y_{2} \end{array} \right) +\left( \begin{array}{c} \varphi \\ 0 \end{array} \right) =A\left( \begin{array}{c} y_{1} \\ y_{2} \end{array} \right) +\left( \begin{array}{c} \varphi \\ 0 \end{array} \right) $$ so $$ \left( \begin{array}{c} y_{1}(x) \\ y_{2}(x) \end{array} \right) =\exp [Ax]\left( \begin{array}{c} y_{1}(0) \\ y_{2}(0) \end{array} \right) +\int_{0}^{x}dy\exp [A(x-y)]\left( \begin{array}{c} \varphi (y) \\ 0 \end{array} \right) $$