Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
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 $y=y_1\,v$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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