Repeated roots · Differential Equations

Intuition

When the characteristic equation has a repeated root, the exponential guess gives only one solution, and a second-order equation needs two. The second is the same exponential multiplied by x. It looks like a trick at first and is nothing of the kind: at a repeated root both the polynomial and its derivative vanish, and those are exactly the two things the new solution needs.

Two exponential rates drifting together until they coincide. Their difference quotient, the difference of the two exponentials divided by the difference of the rates, is a solution all the way, and at the moment of coincidence it becomes the derivative of the exponential with respect to its rate — which is x times the exponential.

Solutions of $y''+2y'+y=0$ , whose characteristic equation has the double root $-1$ . Every solution is $(C_{1}+C_{2}x)e^{-x}$ : it may rise first or cross zero once, and the exponential always wins in the end.

The second solution

If $ar^{2}+br+c=0$ has the repeated root $r=-\tfrac{b}{2a}$ , the general solution of is . The function is a solution because a repeated root is a root of the characteristic polynomial and of its derivative, and it is independent of because their ratio is not constant.

r_{1}=r_{2}=r\ \Longrightarrow\ y=(C_{1}+C_{2}x)\,e^{rx}

Cases and checks

$y''+2y'+y=0$ : $(r$ , so .

Why the second solution carries a factor of x

\begin{aligned}&L\bigl[xe^{rx}\bigr]=p(r)\,xe^{rx}+p'(r)\,e^{rx}\\&p(r)=ar^{2}+br+c\end{aligned}

Differentiate x times the exponential once and twice by the product rule and substitute. Collecting terms, the multiples of x times the exponential add up to the characteristic polynomial at r, and the remaining multiples of the exponential add up to the derivative of that polynomial at r. At a repeated root both vanish, so x times the exponential is a solution.

Proof steps

(xe^{rx})'=(1+rx)\,e^{rx}

The product rule, once.

(xe^{rx})''=(2r+r^{2}x)\,e^{rx}

The product rule again.

L[xe^{rx}]=(ar^{2}+br+c)\,xe^{rx}+(2ar+b)\,e^{rx}

Substitute into a y double prime plus b y prime plus c y and collect the terms.

ar^{2}+br+c=0,\qquad 2ar+b=0

At a repeated root the polynomial and its derivative both vanish.

L[xe^{rx}]=0

So x times the exponential is a second solution.

Applications

Practice

A Repeated Root Gives Two Solutions

One exponential is not enough for a second-order equation; the second solution is $x$ times it.

(r-r_{0})^{2}=0\ \Longrightarrow\ y=(C_{1}+C_{2}x)\,e^{r_{0}x}

Try it

What is the general solution of $y''-6y'+9y=0$ ?