Worked First-Order Initial Value Problems
Chapter Seventy-Six
Syllabus topic Module 1, "1.3 Inverse Laplace Transform"
Pages 168 to 169 of 303
In one line
Five first-order problems worked end to end, every answer substituted back into its own equation by machine.
How to read this chapter
Each problem is worked in the five steps of the previous chapter, and then the answer is put back into the equation. That last step is what a student should do in the examination too: it costs one line and it catches every sign error.
One: a decaying exponential
Solve y prime plus 5y equals 0, y(0) = 3.
Transform: sY minus 3 plus 5Y equals 0. Collect: Y equals 3 over (s plus 5).
L⁻¹{3/(s + 5)} = 3 e^(-5t)
dy/dx + 5y = 0
y = 3 e^(-5x)
Two: a constant forcing term
Solve y prime plus 2y equals 6, y(0) = 1.
Transform: sY minus 1 plus 2Y equals 6 over s. Collect: Y(s plus 2) equals 6 over s plus 1, so Y equals (6 plus s) over s(s plus 2).
Partial fractions, by the cover-up rule: at s = 0 the constant is 6 over 2, which is 3; at s = minus 2 it is 4 over minus 2, which is minus 2.
(s + 6)/(s(s + 2)) = 3/s - 2/(s + 2)
L⁻¹{(s + 6)/(s(s + 2))} = 3 - 2 e^(-2t)
dy/dx + 2y = 6
y = 3 - 2 e^(-2x)
At x = 0 that gives 3 minus 2, which is 1, matching the initial condition. And as x grows it settles at 3, which the final value theorem would have told you without any of the work: s times Y at s = 0 is 6 over 2, which is 3.
lim(s (s + 6)/(s(s + 2)), s -> 0) = 3
Three: an exponential forcing term
Solve y prime minus y equals e to the 2t, y(0) = 0.
Transform: sY minus 0 minus Y equals one over (s minus 2). Collect: Y equals one over (s minus 1)(s minus 2).
1/((s - 1)(s - 2)) = -1/(s - 1) + 1/(s - 2)
L⁻¹{1/((s - 1)(s - 2))} = e^(2t) - e^t
dy/dx - y = e^(2x)
y = e^(2x) - e^x
Four: the resonant case, where the forcing matches the system
Solve y prime minus 2y equals e to the 2t, y(0) = 0.
The forcing term now has the same growth rate as the equation's own solution. Transform: sY minus 2Y equals one over (s minus 2), so Y equals one over (s minus 2) squared.
That is a repeated factor, and the repeated-factors chapter said what that means: a t in the answer.
L⁻¹{1/(s - 2)^2} = t e^(2t)
dy/dx - 2y = e^(2x)
y = x e^(2x)
Worked First-Order Initial Value Problems
The t multiplying the exponential is the signature of resonance, and Module 2 meets exactly the same thing when the exponential rule for a particular integral fails. Two different methods, the same physical fact.
Five: a sinusoidal forcing term
Solve y prime plus y equals sin t, y(0) = 0.
Transform: sY plus Y equals one over (s squared plus 1). Collect: Y equals one over (s plus 1)(s squared plus 1).
Partial fractions with an irreducible quadratic: the term over (s plus 1) has constant one half by covering up, and the quadratic's numerator works out to (1 minus s) over 2.
1/((s + 1)(s^2 + 1)) = 1/(2(s + 1)) + (1 - s)/(2(s^2 + 1))
L⁻¹{1/((s + 1)(s^2 + 1))} = e^(-t)/2 + sin(t)/2 - cos(t)/2
dy/dx + y = sin(x)
y = e^(-x)/2 + sin(x)/2 - cos(x)/2
At x = 0 that is one half plus 0 minus one half, which is 0, matching.
The answer has two parts worth naming. The e to the minus t dies away and is the transient. The sine and cosine persist and are the steady state. That split is what an engineer cares about, and it fell out of the partial fractions by itself: the transient came from the (s plus 1) factor, which is the system, and the steady state from the (s squared plus 1) factor, which is the input.
The pattern across all five
| Denominator of Y | Answer contains |
|---|---|
| a distinct linear factor | an exponential |
| a repeated linear factor | t times an exponential |
| s on its own | a constant |
| an irreducible quadratic | a sine and a cosine |
That is the same table as the partial-fractions chapters, and it means you can predict the shape of a solution before doing any algebra. Predicting the shape first is a good habit, because a solution of the wrong shape is then obvious.
Check yourself
dy/dx + 4y = 0
y = 5 e^(-4x)
dy/dx - 3y = 6
y = 2 e^(3x) - 2
dy/dx + y = e^(-x)
y = x e^(-x)
The third is the resonant case again, in its decaying form: the forcing term is exactly the system's own solution, and the answer carries an x.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.