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:

  1. is this way easier (in your opinion)?
  2. if so, why do we not teach people to do this the easier way? And,
  3. 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 $a=b$.

  • $\begingroup$ This is especially easy in your example because the coefficients are on a particular form : $a+b$ and $ab$. So, it is of very limited range of use. In case of more general form of constant coefficients, this is equivalent to a particular case of the "Reduction of order" thechnique : en.wikipedia.org/wiki/Reduction_of_order . Teaching the general method is more fruitful for a larger range of applications. $\endgroup$ – JJacquelin Sep 8 '15 at 7:38
  • $\begingroup$ Hi @JJacquelin, I'd believe this method is different from the reduction of order method. When $a=b$ reduction of order asks us to guess solutions of the form $v(x) e^{-ax}$. This method requires no guessing. $\endgroup$ – Daniel Mansfield Sep 9 '15 at 2:41
  • $\begingroup$ Also having the coefficients in the form $a+b$ and $ab$ is more to save us the trouble of solving the characteristic equation than a specialisation of the general constant coefficient case. So I don't see the loss of generality here. $\endgroup$ – Daniel Mansfield Sep 9 '15 at 2:44

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 $$ \left( \begin{array}{c} y_{1}(x) \\ y_{2}(x) \end{array} \right) =\exp [Ax]\left( \begin{array}{c} y_{1}(0) \\ y_{2}(0) \end{array} \right) +\int_{0}^{x}dy\exp [A(x-y)]\left( \begin{array}{c} \varphi (y) \\ 0 \end{array} \right) $$

  • $\begingroup$ Interesting, but that does not answer any of the three questions stated by OP. $\endgroup$ – mickep Sep 8 '15 at 8:18
  • $\begingroup$ The first question is about an opinion which is not allowed. So I stated that it is my preferred method, a fact rather than an opinion. The method I outlined is in part in response to Jacquelin's remark about the OP's case being a special one. I do not know why it is not teached nor do I know if it has a specific name. $\endgroup$ – Urgje Sep 8 '15 at 8:27

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