# Question on reduction of order for linear ODEs

If the characteristic equation for a differential equation can be written as $(s-r_1)(s-r_2)$, the substition $z=y'-r_1y$ yields an equation of the form $z'-r_2z=f(x)$.

For example, if our equation is

$y''-3y'+2y=e^x$,

the substitution

$z=y'-2y$

simplifies to

$y''-2y'-(y'-2y)=z'-z=e^x$

At this point, integrating factors can be used to solve for $z$, then substituting back will yield a solution for $y$. My first question is why does it work like this?

My second question is if there's a way to find a substitution for a general second order linear ODE

$y''+p(x)y'+q(x)y=f(x)$

that will similarly reduce the problem to a first order linear ODE which can then be solved by integrating factors?

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Consider the following:

\begin{align} e^{-P}\frac{d}{dx} e^{P-Q} \frac{d}{dx} e^Q y & = e^{-P} \frac{d}{dx} e^P (y' + yQ') \\ &= P'(y' + y Q') + (y'' + y' Q' + y Q'') \\ &= y'' + y' (P' + Q') + y (P'Q' + Q'') \end{align}

which is the reverse of the usual integrating factors procedure. I hope you accept that you can solve $e^{-P}\frac{d}{dx} e^{P-Q} \frac{d}{dx} e^Q y = f$ by algebraic manipulations and repeated integration (the $\frac{d}{dx}$ applies, by convention, to all the things after it).

In the constant coefficient case, if the characteristic equation factors, one can solve for $P$ and $Q$ being linear functions (so $Q'' = 0$). In the variable coefficient case, you end up trying to solve (now writing $u = P'$ and $v = Q'$)

$$u + v = p \qquad v' = q - uv$$

which can be transformed to solving the first order nonlinear ODE

$$v' = q - pv + v^2$$

whose solution may or may not be forthcoming. If you can find an explicit solution to the above ODE, then you can solve for $u,v$ in terms of $p,q,x$ and then integrate to get $P,Q$ in terms of $p,q,x$ and so on and so forth. Note that if $p$ and $q$ are constants, and the polynomial in the right hand side of the above equation has a root, then setting $v$ to be constant equal to that root is a solution of the nonlinear ODE.

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Sorry for taking so long to respond. I wanted to wait until I could focus to go through this. So my substitution in this case is $z=y'+vy$? –  Mike Feb 6 '12 at 13:33
Yes. But the point is more that the "substitution" is just a crutch: the idea behind the "factorisation" is that you can (in some cases) solve your equation using repeated integration with integrating factors. Solving for the functions $P,Q$ (and hence $v$) is the hard part; representing the equation with a substitution $z = y' + vy$ once you know the function $v$ is an exercise in integration by parts. –  Willie Wong Feb 6 '12 at 15:13

First question. Let $D$ be operator thta takes a finction $y$ into its derivative $y'$. The general second order linear equation with constant coefficients can be written as $$(D^2+P\,D+Q)y=f(x),\quad P,Q\in\mathbb{R}.$$ If its characteristic equation is $(s-r_1)(s-r_2)$, then the equation can be factorized as $$(D-r_2)(D-r_1)y=f.$$ The sustitution $z=y'-r_1=(D-r_1)y$ leads to $(D-r_2)z=f$.

Second question. In general, you cannot expect to find easely a factorization of the form $$D^2+P(x)\,D+Q(x)=(D-r_2(x))(D-r_1(x)).$$ There is a procedure called reduction of order to solve a homogeneous linear equation $$y''+P\,y'+Q=0.$$ If you know a solution $y_1$, you can find a second one, linearly independent, with the change $y=y_1\,v$.

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