2
$\begingroup$

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 $a=b$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.