Laplace Inversion and Initial-Value Problems

2 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

The idea

Laplace methods move differentiation into algebra while retaining initial data. The difficult step is often inversion: rewrite the rational expression into known transform pairs, using partial fractions or completing a square.

Visual guide

VISUAL GUIDEInitial data select a damped oscillation

e⁻ᵗ sin t
Upper envelope
Lower envelope

The transform 1/((s + 1)² + 1) inverts to e⁻ᵗ sin t. The exponential envelope bounds the oscillation, while the graph starts at zero with slope one, matching the initial-value problem.

Method and assumptions

Transform both sides, substitute every initial value, solve for $Y(s)$ , decompose and invert term by term. Finally check the initial data and at least one substitution in the original equation. A correct algebraic transform can still be inverted with an incorrect numerator.

Worked example: first-order response

For $y'+2y=3$ , $y(0)=1$ , $(s+2)Y=1+3/s$ . Thus $Y=3/(2s)-1/[2(s+2)]$ and $y=3/2-e^{-2t}/2$ . At $t=0$ it equals $1$ and its steady value is $3/2$ .

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Which time function has transform 1/(s+2)?

e⁻²ᵗ

e²ᵗ

2e⁻ᵗ

Hint 1 · Find a starting point

Use ℒ{eᵃᵗ}=1/(s−a).

Hint 2 · Take the next step

Here a=−2.

Show the reasoning

Answer: e⁻²ᵗ

The inverse is e⁻²ᵗ for t≥0, with transform defined for Re(s)>−2.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: damped oscillation

For $y''+2y'+2y=0$ , $y(0)=0,y'(0)=1$ , $Y=1/(s^2+2s+2)=1/[(s+1)^2+1]$ . The inverse is $e^{-t}\sin t$ , which has the required initial slope.

Interpreting the result

A shift in $s$ produces an exponential in time. A factor $e^{-as}$ produces a time delay instead; these are different transform rules. Repeated poles require terms for each power in partial fractions.

Practice

Invert $1/[s(s+1)]$ .
Invert $1/(s+2)^2$ .
Invert $(s+1)/[(s+1)^2+4]$ .

Show worked solutions

$1/s-1/(s+1)$ gives $1-e^{-t}$ .
$te^{-2t}$ .
$e^{-t}\cos2t$ .

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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