# Method of Undetermined Coefficients for Higher Order Linear Equations

1. May 5, 2013

### craighenn

Hi, I'd just like to have a quick clarification with regards to the method of undetermined coefficients. I know that if a characteristic equation has the form

(r-4)3 = 0

then the characteristic solution will be

yc = e4t + te4t + t2e4t + t3e4t
and the particular solution ought to be

Y = At4e4t

That all makes sense to me. But I was wondering how the particular solution is to be found if the characteristic solution isn't entirely one root. Like if it was

(r-4)2(r-2) =0

The characteristic solution would be

yc = e4t + te4t + e2t

My question then is two-fold. First, how do I structure the particular solution with regard to the method of undetermined coefficients? Because when all the roots repeat then the particular is just one power of t greater than the greatest one in the solution, like in my first example.

But with two different roots out of 3 (or think of any other example), I'm not sure what to do. Can I even use that method, or is a particular solution even necessary when the characteristic equation has more than just repeated roots?

Hopefully I wrote this clear enough to convey what I'm asking.

2. May 6, 2013

### HallsofIvy

The only time a characteristic root is important in determining a particular solution is when it is one of the possiblities for the particular solution. So if your characteristic sollution is $C_1e^{4t}+ C_2te^{4t}+ C_3e^{2t}$ (don't forget the undetermined constants) then whether you need to consider them at all depends on the "right hand side". If the right hand side were $e^{5t}$ and does not involve any of the characteristic roots, then you would just try $Ae^{5t}$. If there were a "$e^{4t}$" term, since you already have $e^{4t}$ and $te^{4t}$, you would need to try $t^2e^{4t}$. The important point is that in determining whether to "multiply t" is that you only have to determine whether that particular exponent is one of the characteristic roots.

Here's another question you may not have thought about- suppose the right hand side already involves a power of t along with an exponential. Say, the differential equation is $y''- 6y'+ 9y= t^2e^{3t}$ The characteristic equation is $r^2- 6r+ 9= (r- 3)^2= 0$ and has r= 3 as a double root. That means the characteristic solution is of the form $C_1e^{3t}+ C_2te^{3t}$. Normally, with $t^2e^{3t}$ on the right, we would try $(At^2+ Bt+ C)e^{3t}$- that is we want all powers of t from the $t^2$ we have on the right, down. But now because we already have $e^{3t}$ and $te^{3t}$, we multiply that by $t^2$ and try $(At^4+ Bt^3+ Ct^2)e^{3t}$.

Last edited by a moderator: May 6, 2013