**Background for students**

Remember that when one is trying to solve a non-homogeneous differential equation, say:

one finds the general solution to (which is called the homogeneous solution; in this case it is and then finds some solution to . This solution, called a particular solution, will not have an arbitrary constant. Hence that solution cannot meet an arbitrary initial condition.

But adding the homogenous solution to the particular solution yields a general solution with arbitrary constants which can be solved for to meet a given initial condition.

So how does one obtain a particular solution?

Students almost always learn the so-called “method of undetermined coefficients”; this is used when the driving function is a sine, cosine, , a polynomial, or some sum and product of such things. Basically, one assumes that the particular solution has a certain form than then substitutes into the differential equation and then determines the coefficients. For example, in our example, one might try and then substitute into the differential equation to solve for and . One could also try a complex form; that is, try and then determines and then uses the real part of the solution.

A second method for finding particular solution is to use variation of parameters. Here is how that goes: one obtains two linearly independent homogeneous solutions and then seeks a particular solution of the form where and where is the determinant of the Wronskian matrix. This method can solve differential equations like and sometimes is easier to use when the driving function is messy.

But sometimes it can lead to messy, non transparent solutions when “undetermined coefficients” is much easier; for example, try solving with variation of parameters. Then try to do it with undetermined coefficients; though the answers are the same, one method yields a far “cleaner” answer.

**There is a third way that gives a particular solution** that meets a specific initial condition. Though this method can yield a not-so-easy-to-do-by-hand integral and can sometimes lead to what I might call an answer in obscured form, the answer is in the form of a definite integral that can be evaluated by numerical integration techniques (if one wants, say, the graph of a solution).

This method is the **Convolution Method**. Many texts introduce convolutions in the Laplace transform section but there is no need to wait until then.

**What is a convolution?**

We can define the convolution of two functions and to be:

. Needless to say, and need to meet appropriate “integrability” conditions; this is usually not a problem in a differential equations course.

Example: if , then . Notice that the dummy variable gets “integrated out” and the variable remains.

There are many properties of convolutions that I won’t get into here; one interesting one is that ; proving this is an interesting exercise in change of variable techniques in integration.

**The Convolution Method**

If is a homogenous solution to a second order linear differential equation that meets initial conditions: and is the forcing function, then is the particular solution that meets

How might we use this method and why is it true? We’ll answer the “how” question first.

Suppose we want to solve . The homogeneous solution is and it is easy to see that we need to meet the condition. So a particular solution is

This particular solution meets .

**Why does this work?**

This is where “differentiation under the integral sign” comes into play. So we write .

Then ?

Look at the convolution integral as . Now think of . Then from calculus III: . Of course, .

by the Fundamental Theorem of calculus and by differentiation under the integral sign.

So we let and we see which equals because . Now by the same reasoning because .

Now substitute into the differential equation and use the linear property of integrals to obtain

It is easy to see that Now check .

“But adding the particular solution to the particular solution yields a general solution with arbitrary constants which can be solved for to meet a given initial condition. ”

I believe you mean adding the “… homogeneous solution to the particular solution …”

Good article, though.

Comment by Josh — September 28, 2012 @ 7:59 pm

Thanks Josh! I made the correction.

Comment by collegemathteaching — September 28, 2012 @ 9:23 pm

[…] Of course, the convolution method has utility beyond the Laplace transform setting. […]

Pingback by Why I teach multiple methods for the inverse Laplace Transform. | College Math Teaching — November 12, 2013 @ 8:33 pm

[…] I’ve discussed a type of convolution integral with regards to solving differential equations […]

Pingback by The convolution integral: do some examples in Calculus III or not? | College Math Teaching — August 31, 2014 @ 11:32 pm

Elequently and acurate, thank you very much

Matia Yoav, P.hD

Comment by yoav matia — June 23, 2015 @ 11:26 am

sorry for the erroneous spelling check:

“Eloquently and accurate, thank you very much”

Comment by yoav matia — June 23, 2015 @ 11:29 am

I believe sin(t)*tan(t) = tan(t)*sin(t) is wrong as it should be sin(t)*tan(t) = -tan(t)*sin(t)

Comment by Diego — November 28, 2015 @ 9:27 pm