Ask Uncle Colin: Auxiliary Equations With Repeated Roots

In this post, I swap liberally between d-notation and '-notation for derivatives. Deal with it.

Dear Uncle Colin,

Why do we have to treat second-order ODEs differently when the auxiliary equation has a repeated root?

• Something Or Other Defies Expectations

Hi, SOODE, and thanks for your message!

First, some background for everyone else.

If you have a second-order ordinary differential equation such as $\diffn{2}{y}{x} + 3\dydx + 2y = 0$ [equation 1], one method is to pull out the auxiliary equation, $D^2 + 3D + 2 = 0$, which has the two roots $D=-2$ and $D=-1$.

These correspond to two solutions to the ODE, $y=e^{-2x}$ and $y=e^{-x}$ - the general solution being any linear combination of the two, $Ae^{-2x} + Be^{-x}$.

That's good, and extends nicely into the complex numbers once you're ok with the exponential forms of sine and cosine.

However, there's a glitch

If you have something like $\diffn{2}{y}{x} + 4\dydx + 4y = 0$ [equation 2], the auxiliary equation is $D^2 + 4D + 4 = 0$, which has a repeated root. It turns out that the general solution is $y=(Ax+B)e^{-2x}$, which hints at a completely different method.

In a sense, yes. In another, no.

Where all this comes from

Let's dial back a bit and consider solving [equation 1] without magical shortcuts, instead using a substitution: let $y = Y e^{kx}$.

We get $y^\prime = Y^\prime e^{kx} + k Y e^{kx}$ and $y^{\prime\prime} = Y^{\prime\prime} e^{kx} + 2k Y^\prime e^{kx} + k^2 Y e^{kx}$.

This transforms the original equation into $e^{kx} \br{Y^{\prime\prime} + (2k+3)Y^\prime + (k^2 + 3k + 2)Y}=0$, for whatever value of $k$ we pick - notice that the $Y$ coefficient is the left-hand side of the auxiliary equation.

If we pick $k$ to make the $Y$ term vanish, it makes things rather simpler - and it doesn't matter which we pick. Let's go for $k=-1$, which makes the equation:

$e^{-x}\br{Y^{\prime\prime} + Y^\prime} = 0$ (and clearly, the $e^{-x}$ term can be divided out as it's never zero.)

Integrating with respect to $x$, we get $Y^\prime + Y = A$, which we can solve with an integrating factor to find $\diff{\br{Ye^{x}}}{x} = Ae^{x}$, so $Ye^{x} = Ae^{x} + B$, or $Y= A + Be^{-x}$.

Reversing the original substitution gives $y = Ae^{-x} + Be^{-2x}$, as previously.

And with repeated roots?

If we start with $y^{\prime\prime} + 4y^\prime + 4y = 0$ and use the same substitution as before, we find it transforms everything into $e^{kx} \br{Y^{\prime\prime} + (2k + 4) Y^\prime + (k^2 + 4k + 4)Y}=0$.

Again, we pick $k$ to eliminate the $Y$ term - which has the side-effect of eliminating the $Y^\prime$ term, too! If $k=-2$, we have $Y^{\prime\prime} = 0$, after ignoring the exponential factor as before.

This integrates to $Y^\prime = A$ and $Y = Ax + B$; reversing the substitution gives $y=(Ax+B)e^{-2x}$.

This, of course, is an explanation rather than a proof; a proof would follow roughly the same steps, but involve rather more letters.

Hope that helps!

- Uncle Colin

* Edited 2018-05-02 to correct a mix-up of $x$s and $t$s. Thanks to @robjlow for pointing out the problem!

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

Share

This site uses Akismet to reduce spam. Learn how your comment data is processed.