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
using the Laplace transform approach.
Although it is not stated explicitly we shall assume that is a causal function (we have no interest in the value of if .) Similarly, the function on the right-hand side of the differential equation , the ‘forcing function’, will be assumed to be causal. (Strictly, we should write but the step function will often be omitted.) Let us write . Then, taking the Laplace transform of every term in the differential equation gives:
Now
Substituting these expressions into the transformed version of the differential equation gives:
Solving for we have
Therefore
Now, using partial fractions, this last expression can be written in a more convenient form:
and then, inverting:
thus
This is the solution to the given initial value problem.
Task!
The equation governing the build up of charge, , on the capacitor of an RC circuit is
where is the constant d.c. voltage. Initially, the circuit is relaxed and the circuit is then ‘closed’ at and so is the initial condition for the charge.
Use the Laplace transform method to solve the differential equation for .
Assume the forcing term is causal.
Begin by finding an expression for :
since, taking the Laplace transform of each term in the differential equation:
where, we emphasize, the Laplace transform of the constant term is .
Inserting we have, after some rearrangement,
fractions:
You should obtain Now obtain by taking inverse Laplace transforms:
since
The solution to this problem is illustrated in the following diagram.
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:
using the Laplace transform method.
Solution
As usual we shall assume the forcing function is causal (i.e. is really .0 Taking the Laplace transform of each term:
that is,
Inserting the initial conditions and rearranging:
Then, using partial fractions:
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:
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 and the frictional force divided by this mass is .
- Write down the equation governing the motion in terms of the displacement m and time seconds including the impulse .
- Write down the initial conditions on the displacement and velocity.
- Solve the equation for displacement as a function of time.
- Draw a graph of the oscillations for to 10 s.
Solution
-
Since the system involves a restoring force and friction, after dividing through by the mass, the equation of motion may be written:
where the right-hand side represents the impulse being switched on at s and switched off at s.
- Since the system starts from rest .
-
Taking the Laplace Transform of each term of the differential equation gives
i.e.
but as , this simplifies to
so, on taking inverse Laplace Transforms,
-
Figure 16
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.