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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?

2 Answers 2

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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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    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.2012-02-06
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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 $y=y_1\,v$.