Skip to the content of the web site.

Lecture 15

Lecture 14 | Lecture 16


Applying the Laplace Transform to Differential Equations

The Laplace transform may be used to simplify the solving of linear differential equations with constant coefficients. Starting with a very simple example, suppose we have the differential equation

tex:$$\ddot{x} + 3\dot{x} + 2x = 0$$
tex:$$x(0) = 1$$
tex:$$\dot{x}(0) = 0$$.

To solve this using the Laplace transform, assume tex:$$x(t) \leftrightarrow X(s)$$ we take the Laplace transform of both sides:

tex:$$L\left\{ \ddot{x} + 3\dot{x} + 2x \right\} = L\{0\}$$

or, by linearity

tex:$$L\left\{ \ddot{x} \right\} + 3 L\left\{ \dot{x} \right\} + 2 L\left\{x\right\} = 0$$

and by using the rule for the Laplace transform of derivatives:

tex:$$[s^2 X(s) - sx(0) - \dot{x}(0)] + 3[s X(s) - x(0)] + 2X(s) = 0$$.

By using the initial conditions, we get:

tex:$$s^2 X(s) - s + 3s X(s) - 3 + 2X(s) = (s^2 + 3s + 2)X(s) - (s + 3) = 0$$.

By solving this equation for tex:$$X(s)$$, we get that

tex:$$X(s) = \frac{s + 3}{s^2 + 3s + 2}$$.

The question is, what function tex:$$x(t)$$ has tex:$$\frac{s + 3}{s^2 + 3s + 2}$$ as its Laplace transform? This process is called finding the inverse Laplace transform of tex:$$\frac{s + 3}{s^2 + 3s + 2}$$.

In this example, we can use partial fraction decomposition to aid us: by observing that tex:$$s^2 + 3s + 2 = (s + 1)(s + 2)$$, and therefore tex:$$\frac{s + 3}{(s + 1)(s + 2)} = \frac{a}{s + 1} + \frac{b}{s + 2}$$. Using the cover-up method, one may quickly determine that tex:$$\frac{s + 3}{(s + 1)(s + 2)} = \frac{2}{s + 1} - \frac{1}{s + 2}$$.

To do this in Maple, use the command:

> convert( (s + 3)/(s^2 + 3*s + 2), parfrac, s );

However, by inspection, we observe that tex:$$e^{-t} \leftrightarrow \frac{1}{s + 1}$$ and tex:$$e^{-2t} \leftrightarrow \frac{1}{s + 2}$$; therefore tex:$$2e^{-t} - e^{-2t} \leftrightarrow \frac{s + 3}{(s + 1)(s + 2)}$$.

Consequently, tex:$$x(t) = 2e^{-t} - e^{-2t}$$.

Remark 1

The above example is a specific example of the more general 2nd-order differential equation

tex:$$a_2 \ddot{x} + a_1 \dot{x} + a_0 = f(t)$$
tex:$$x(0) = x_0$$
tex:$$\dot{x}(0) = \dot{x}_0$$

where all of the parameters are given parameters. In a mechanical spring system, the parameters are tex:$$m$$, tex:$$\gamma$$, and tex:$$k$$ while in an electrical system, the parameters are tex:$$L$$, tex:$$R$$, and tex:$$\frac{1}{C}$$.

The function tex:$$f(t)$$ is the forcing function and the solution tex:$$x(t)$$ is called the response of the system.

Remark 2

The same situation applies to systems of 1st-order linear differential equations with constant coefficients. See AMEM, p.126-8 to see the solution to the system

tex:$$\dot{x} + \dot{y} + 5x + 3y = e^{-t}$$
tex:$$2\dot{x} + \dot{y} + x + y = 3$$

where tex:$$x(0) = 2$$ and tex:$$y(0) = 1$$.