Laplace Transform · Lesson

Topics covered

Directly verify that the solution obtained by transforming a time-domain differential equation satisfies the original problem.

Overall workflow

Differential equation and initial values → s-domain equation → Y(s) → time response and residuals

Examining expressions and their domains

Canceling common factors in a rational expression simplifies calculations. However, values that make the original denominator 0 must still be excluded. For example, $\frac{x-1}{x-1}$ equals 1 at $x\ne1$, but at $x=1$ the original expression is undefined.

Converting a differential equation to an algebraic equation

Applying the Laplace transform to $\dot y+2y=1$ separates the effects of input and initial state into distinct terms.

$$(s+2)Y(s)=\frac{1}{s}+y(0)$$

The two solutions for different initial values are shown below.

$$y(t)=\frac12+\left(y(0)-\frac12\right)e^{-2t}$$

Substituting y(0)=0 and 1 gives the same steady-state value but different starting points and transient responses.

Two responses with different initial values

For $\dot y+2y=1$, compare y(0)=0 and 1. Plot both solutions on the same time axis and check their initial and steady-state values.

import numpy as np
import matplotlib.pyplot as plt
t=np.linspace(0,4,401)
for initial in [0,1]:
    y=.5+(initial-.5)*np.exp(-2*t)
    plt.plot(t,y,label=f'y(0)={initial}')
plt.axhline(.5,color='gray',linestyle='--');plt.xlabel('Time [s]');plt.ylabel('y');plt.legend();plt.grid()

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