1 Solving ODEs using Laplace transforms

We begin with a straightforward initial value problem involving a first order constant coefficient differential equation. Let us find the solution of

$\frac{dy}{dt}+2y=12{\text{e}}^{3t}y\left(0\right)=3$

using the Laplace transform approach.

Although it is not stated explicitly we shall assume that $y\left(t\right)$ is a causal function (we have no interest in the value of $y\left(t\right)$ if $t<0$ .) Similarly, the function on the right-hand side of the differential equation $\left(12{\text{e}}^{3t}\right)$ , the ‘forcing function’, will be assumed to be causal. (Strictly, we should write $12{\text{e}}^{3t}u\left(t\right)$ but the step function $u\left(t\right)$ will often be omitted.) Let us write $\mathcal{L}\left\{y\left(t\right)\right\}=Y\left(s\right)$ . Then, taking the Laplace transform of every term in the differential equation gives:

$\mathcal{L}\left\{\frac{dy}{dt}\right\}+\mathcal{L}\left\{2y\right\}=\mathcal{L}\left\{12{\text{e}}^{3t}\right\}$

Now

Substituting these expressions into the transformed version of the differential equation gives:

$\left[-3+sY\left(s\right)\right]+2Y\left(s\right)=\frac{12}{s-3}$

Solving for $Y\left(s\right)$ we have

$\left(s+2\right)Y\left(s\right)=\frac{12}{s-3}+3=\frac{3+3s}{s-3}$

Therefore

$Y\left(s\right)=\frac{3\left(s+1\right)}{\left(s+2\right)\left(s-3\right)}$

Now, using partial fractions, this last expression can be written in a more convenient form:

$Y\left(s\right)=\frac{3∕5}{\left(s+2\right)}+\frac{12∕5}{\left(s-3\right)}$

and then, inverting:

$y\left(t\right)={\mathcal{L}}^{-1}\left\{Y\left(s\right)\right\}=\frac{3}{5}{\mathcal{L}}^{-1}\left\{\frac{1}{s+2}\right\}+\frac{12}{5}{\mathcal{L}}^{-1}\left\{\frac{1}{s-3}\right\}$

thus

$y\left(t\right)=\frac{3}{5}{\text{e}}^{-2t}u\left(t\right)+\frac{12}{5}{\text{e}}^{3t}u\left(t\right)$

This is the solution to the given initial value problem.

Task!

The equation governing the build up of charge, $q\left(t\right)$ , on the capacitor of an RC circuit is $R\frac{dq}{dt}+\frac{1}{C}q=v_0$

where $v_0$ is the constant d.c. voltage. Initially, the circuit is relaxed and the circuit is then ‘closed’ at $t=0$ and so $q\left(0\right)=0$ is the initial condition for the charge.

Use the Laplace transform method to solve the differential equation for $q\left(t\right)$ .

Assume the forcing term $v_0$ is causal.

Begin by finding an expression for $Q\left(s\right)=\mathcal{L}\left\{q\left(t\right)\right\}$ :

Answer

fractions:

Answer

Answer

The Laplace transform method is also applied to higher-order differential equations in a similar way.

Example 1

Solve the second-order initial-value problem:

$\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}+2y={\text{e}}^{-t}y\left(0\right)=0,y^{\prime }\left(0\right)=0$

using the Laplace transform method.

Solution

As usual we shall assume the forcing function is causal (i.e. is really ${\text{e}}^{-t}u\left(t\right)$ .0 Taking the Laplace transform of each term:

$\mathcal{L}\left\{\frac{d^{2}y}{d{t}^2}\right\}+2\mathcal{L}\left\{\frac{dy}{dt}\right\}+2\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{{\text{e}}^{-t}\right\}$

that is,

$\left[-y^{\prime }\left(0\right)-sy\left(0\right)+s^{2}Y\left(s\right)\right]+2\left[-y\left(0\right)+sY\left(s\right)\right]+2Y\left(s\right)=\frac{1}{s+1}$

Inserting the initial conditions and rearranging:

$Y\left(s\right)\left[s^2+2s+2\right]=\frac{1}{s+1}\text{ i.e.}Y\left(s\right)=\frac{1}{\left(s+1\right)\left(s^2+2s+2\right)}$

Then, using partial fractions:

$\frac{1}{\left(s+1\right)\left(s^2+2s+2\right)}\equiv \frac{1}{s+1}-\frac{\left(s+1\right)}{s^2+2s+2}\equiv \frac{1}{s+1}-\frac{\left(s+1\right)}{{\left(s+1\right)}^2+1}$

where we have completed the square in the second term of the right-hand side. We can now take the inverse Laplace transform:

which is the solution to the initial value problem.

Exercises

Use Laplace transforms to solve:

1. $\frac{dx}{dt}+x=9{\text{e}}^{2t}x\left(0\right)=3$
2. $\frac{d^{2}x}{d{t}^2}+x=2tx\left(0\right)=0x^{\prime }\left(0\right)=5$

Answer

Example 2

A damped spring, constrained to move in one direction, such as might be found in a railway buffer, is subjected to an impulse of duration 5 seconds. The spring constant divided by the mass causing the impulse is $10{\text{m}}^{-2}{\text{s}}^{-2}$ and the frictional force divided by this mass is $2{\text{m}}^{-2}{\text{s}}^{-2}$ .

1. Write down the equation governing the motion in terms of the displacement $x$ m and time $t$ seconds including the impulse $u\left(t\right)$ .
2. Write down the initial conditions on the displacement $\left(x\right)$ and velocity.
3. Solve the equation for displacement as a function of time.
4. Draw a graph of the oscillations for $t=0$ to 10 s.

Solution

1. Since the system involves a restoring force and friction, after dividing through by the mass, the equation of motion may be written:

   $\frac{d^{2}x}{d{t}^2}+2\frac{dx}{dt}+10x=u\left(t\right)-u\left(t-5\right)$

   where the right-hand side represents the impulse being switched on at $t=0$ s and switched off at $t=5$ s.

2. Since the system starts from rest $x\left(0\right)=x^{\prime }\left(0\right)=0$ .
3. Taking the Laplace Transform of each term of the differential equation gives

   $\mathcal{L}\left[\frac{d^{2}x}{d{t}^2}\right]+2\mathcal{L}\left[\frac{dx}{dt}\right]+10\mathcal{L}\left[x\right]=\mathcal{L}\left[u\left(t\right)\right]-\mathcal{L}\left[u\left(t-5\right)\right]$

   i.e. $s^2\bar{x}\left(s\right)-x\left(0\right)-s{x}^{\prime }\left(0\right)+2\left(s\bar{x}\left(s\right)-x\left(0\right)\right)+10\bar{x}\left(s\right)=\frac{1}{s}-\frac{1}{s}{\text{e}}^{-5s}$

   but as $x\left(0\right)=x^{\prime }\left(0\right)=0$ , this simplifies to $s^2\bar{x}\left(s\right)+2s\bar{x}\left(s\right)+10\bar{x}\left(s\right)=\frac{1}{s}-\frac{1}{s}{\text{e}}^{-as}$

   $$\begin{array}{rcll}\text{i.e.}\bar{x}\left(s\right)& =& \frac{1}{s^2+2s+10}\frac{1}{s}\left[1-{\text{e}}^{-5s}\right]& \\ & =& \left[\frac{1}{10}\frac{1}{s}-\frac{1}{10}\frac{s+2}{s^2+2s+10}\right]\left[1-{\text{e}}^{-5s}\right]& \\ & =& \left[\frac{1}{10}\frac{1}{s}-\frac{1}{10}\frac{s+1}{{\left(s+1\right)}^2+3^2}-\frac{1}{30}\frac{3}{{\left(s+1\right)}^2+3^2}\right]\left[1-{\text{e}}^{-5s}\right]& \\ & =& \frac{1}{10}\frac{1}{s}-\frac{1}{10}\frac{s+1}{{\left(s+1\right)}^2+3^2}-\frac{1}{30}\frac{3}{{\left(s+1\right)}^2+3^2}& \\ & & -\frac{1}{10}\frac{1}{s}{\text{e}}^{-5s}+\frac{1}{10}\frac{s+1}{{\left(s+1\right)}^2+3^2}{\text{e}}^{-5s}+\frac{1}{30}\frac{3}{{\left(s+1\right)}^2+3^2}{\text{e}}^{-5s}& \end{array}$$

   so, on taking inverse Laplace Transforms,

   $$\begin{array}{rcll}x\left(t\right)& =& \frac{1}{10}-\frac{1}{10}{\text{e}}^{-t}cos3t-\frac{1}{30}{\text{e}}^{-t}sin3t& \\ & & -\frac{1}{10}u\left(t-5\right)+\frac{1}{10}{\text{e}}^{-\left(t-5\right)}cos3\left(t-5\right)u\left(t-5\right)+\frac{1}{30}{\text{e}}^{-\left(t-5\right)}sin3\left(t-5\right)u\left(t-5\right)& \end{array}$$
4. Figure 16

   

   Long description

   According to the graph the damped spring has a damped oscillation about a displacement of 0.1 m after the start of the impulse and a damped oscillation about a displacement of zero after the impulse has finished.