Engineering Math Series 10: Solving ODEs with the Laplace Transform

The value of the Laplace transform lies not so much in its definition but in the fact that **initial value problems can be solved through a very consistent procedure**. In this post, let us follow that entire flow by hand.

The Most Important Formulas

Letting $Y(s)$ denote the Laplace transform of $y(t)$,

$$\mathcal{L}\{y'(t)\} = sY(s) - y(0)$$

$$\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)$$

The initial conditions are automatically incorporated into the equation. This is precisely why it is especially convenient for initial value problems.

Standard Procedure

1. Apply the Laplace transform to the entire differential equation

2. Substitute initial conditions to create an algebraic equation for $Y(s)$

3. Solve for $Y(s)$

4. Use partial fraction decomposition and take the inverse Laplace transform

Worked Example

Consider the following initial value problem.

$$y' + y = 1, \quad y(0)=0$$

Applying the Laplace transform to both sides,

$$\mathcal{L}\{y'\} + \mathcal{L}\{y\} = \mathcal{L}\{1\}$$

which gives

$$sY(s) - y(0) + Y(s) = \frac{1}{s}$$

Substituting the initial condition $y(0)=0$,

$$sY(s) + Y(s) = \frac{1}{s}$$

Therefore

$$Y(s)(s+1) = \frac{1}{s}$$

$$Y(s) = \frac{1}{s(s+1)}$$

Performing partial fraction decomposition,

$$\frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}$$

Taking the inverse transform,

$$y(t) = 1 - e^{-t}$$

Interpreting the Result

This solution starts at 0 when time is 0 and gradually converges to 1. It is the typical first-order response where the system exponentially stabilizes toward the external input of 1.

The reason the Laplace transform is great is that this process can be handled through an almost mechanical procedure, without needing to come up with something new each time like the integrating factor method.

Thinking One Step Further

It works the same way for second-order problems too. For example, given

$$y'' + 3y' + 2y = 0$$

with initial conditions, after transformation it becomes a rational expression in $Y(s)$, which eventually reduces to a partial fraction decomposition problem.

In other words, the Laplace transform is less of "a tool that converts differential equations into integration problems" and more like "a tool that converts differential equations into algebra and table-lookup problems" in practice.

Engineering Applications

Step Response

One of the most common inputs in control engineering is the step input. In the Laplace domain, the step input is simply $1/s$, which greatly simplifies the calculation.

Impulse Response

The response of a system to an instantaneous shock can also be handled cleanly within the Laplace transform framework.

Block Diagram Analysis

Block diagrams that connect transfer functions are mostly analyzed from the Laplace domain perspective.

Common Mistakes

Rushing the algebraic simplification after transformation

If you do not carefully organize $Y(s)$ terms and constant terms, sign errors are frequent.

Skipping partial fraction decomposition

The inverse transform usually requires standard forms to be visible. Therefore, the decomposition step is very important.

Finishing without interpretation

If the result is something like $1 - e^{-t}$, you must always develop the habit of interpreting the initial value, final value, and direction of change.

One-Line Summary

The Laplace transform turns linear differential equations with initial conditions into a systematic algebraic procedure.

Next Post Preview

Starting with the next post, we will broaden the language of differential equations and enter **matrices and linear systems** that handle multiple unknowns at once.

References

- Erwin Kreyszig, _Advanced Engineering Mathematics_, 10th Edition

- Alan V. Oppenheim, Alan S. Willsky, _Signals and Systems_

- Dennis G. Zill, _Differential Equations with Boundary-Value Problems_