Use Laplace Transforms to Solve Differential Equations

And here comes the feature of Laplace transforms handy that a derivative in the "t"-space will be just a multiple of the original transform in the "s"-space.

That's precisely what we are going to do:

Apply Laplace Transform to all terms of a D.E.:

Derivatives disappear, multiples of the (transformed) function instead !

The best way to show this is through an example:

Example: 2nd order nonhomogeneous D.E. with initial values:

This should look quite familiar: It is a 2nd order LDE , nonhomogeneous, coefficients are constant … wait a moment - why are there initial conditions?


Answer: well - a little downfall of this method - it is good for solving initial value problems only. But very often we can live with this rather happily - since in professional life we are supposed to solve particular problems - almost nobody is interested in general solutions anymore.

Let's start our procedure (and this is what it is: just a procedure you have to follow step by step, and you will succeed!)

Step 1: Laplace transform all terms of the D.E.:

(Almost) no thinking required here: Take the differential equation - that would be in our example, and apply the theorem to each term of the equation separately (we can do this because the Laplace transform is - linear):

where are the Laplace transform and

This new equation in the "s"-space is called "subsidary Equation"

We can sort terms and see, of course, that in the "s"-space of Laplace transforms, our former Differential equation contains no derivatives anymore: We end up with an embarrassing simple algebraic equation for Y(s).

Note, that y(0) and y'(0) are no functions anymore but just our initial values - this is where our initial values are coming in!

not a D.E. anymore !!

As I said, the next step will be embarrassing easy: We have to solve the subsidary equation for Y(s). Let's do this:

Step 2: Solve the subsidary equation

That’s easy:


or, with (transfer function)

To let this procedure appear tructured (and your life a little more complicated, we can call the denominator of this expression "transfer function" and call it Q(s). Then, we can write our D.E. in just one line such as

Hurrah! Well, it really looks much better now. However, the simplicity of this equation becomes entirely clear when we remember that are just numbers ….


They are actually very special constants ….

Right, they are our initial conditions!

In order to keep the following as simple as possible (don't worry, we can make it arbitrarily complicated later on) we set the initial conditions in this example equal to zero:

In this particular case our transfer function Q turns out to be very simple:

For : and then (but only then)

(but only for these initial conditions - please think about why this is so!)

The transfer function has even a real measurable meaning in applications: It is the ratio between the transforms of the output and input amplitudes of an electric circuit, for example:

Physically this means (e.g. for an electrical circuit) that

Q here describes the transfer of energy put into the system to the energy that leaves the system.

But let's get back to our problem: Although we found a solution for Y(s) we are still in the "s"-space! Our solution, however, is required in the real world of the "t"-space.

The quest now is to transform our solution Y(s) back into the a real solution y(t):

3rd and last step: Use e.g. partial fractions to reduce

into simple terms that can be looked up in tables

This sounds simple but sometimes ends up to be quite a painful process! However in this course, most of the times partial fractions can be found relatively easily.

Did you understand the procedure so far? You might want to go back to the first step and read it again….

However, we also can demonstrate it with another, more particular example:

Let’s do a simple problem:

This is really simple - we got a second order LDE with constant coefficients, a=0, b=1, and our r(t)=t

Step 1: Transforming this equation is not diffficult, we use the definition for the transform of the second derivative for the left side of the equation, and look up the transform for "t":

, therefore

and finally solve it for Y…

Step 2: Solves to

… and rewrite our solution in partial fractions.

Now we have . What is left over to do is to transform back these simple expressions (use a table, for example):

Step 3: Transfer back by table look-up:

And so we proceed, term by term by term. Done. Our solution is

Try for yourself and test this solution for this particular problem. Just to make sure it works!


Alright, once again the entire procedure, this time as a diagram (Start in the upper left corner and follow the arrows):




Subsidary equation


Solution of the given problem:

Solution of subsidary equation


Let's do this again, with a more elaborate D.E.:

Another example:


This time, a=2, and


Laplace transform:

O.K., step 1: we Laplace transform the entire equation (left side using the def's for the first and second derivatives, right side a table look up)


Solve for Y(s) …

… and simplify Y(s) by finding partial fractions ….

Which results into (2nd step)

And then, we transform back, term by term, into the "t"-space:






Now since we have seen how the derivative of a function Laplace transforms, and how beneficial this behavior can be for us in order to solve differential equations, we should consequently pay attention into the opposite process, the integration.

Copyrights 1999, 2000 by Peter Dragovitsch and Ben A. Fusaro