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.

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