Just a brief review: if we have a second order linear differential equation with constant coefficients we obtain the solution in the following manner: first we solve the related homogeneous differential equation to obtain Then, if is the “right” kind of function, one can make a guess at what a particular solution would be up to a constant multiple. One then inserts their candidate for the particular solution into the differential equation to solve for the constant.

What is the “right” kind of function? Basically these are the classes of functions that form a closed class under differentiation. Examples of such closed classes would be exponential functions, sines and cosines (together), polynomials, and possibly multiples of these.

Here is an example to refresh your memory: suppose Then and we assume

Then putting back into the differential equation we obtain which implies So it follows that the general solution is

Trouble arises when the forcing function is itself part of the homogeneous solution; example

We usually tell our students to multiply a candidate for the particular solution by and try again; e. g. in this example we’d attempt which works out to . But when does this work?

Let’s see: Suppose we have and we attempt where and are two linearly independent homogeneous solutions. Put these into the differential equation and do some algebra and we obtain:

Case One:

(note: we can let if needed)

So and then becomes:

Equating

coefficients with the forcing function can lead to a matrix system whose

left hand side is:

The coefficient matrix has rank 2 even if and Hence

the system can always be solved.

Case Two

Then and then becomes:

which is always solvable unless or In either case, this implies that or are themselves part of the homogeneous solution. This puts us in

Case Three:

Here: will NOT work as the first term is part of the homogeneous solution. This is the only case where is required.

So attempt

Putting into the differential equation we end up with

which upon substituting , becomes:

.

But remember that as we were in the “real repeated root situation” so this becomes:

Which, of course, can always be solved. Note that plays no role in the coefficients.

Example: if Then one can easily check that a particular solution is given by

[…] didn’t post much today but I did write this post about the “method of undetermined coefficients” for solving inhomogeneous second order […]

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