Laplace Transforms

Study Sheet

Laplace Transforms

Turning calculus into algebra

The Transform Method

Tip
The transform and the short table

L{f}(s)=∫0∞f(t)e−stdt\mathcal L\{f\}(s) = \int_0^\infty f(t)e^{-st}dt. The working table: L{eat}=1s−a\mathcal L\{e^{at}\} = \tfrac{1}{s-a}, L{tn}=n!sn+1\mathcal L\{t^n\} = \tfrac{n!}{s^{n+1}}, L{sin⁡bt}=bs2+b2\mathcal L\{\sin bt\} = \tfrac{b}{s^2+b^2}, L{cos⁡bt}=ss2+b2\mathcal L\{\cos bt\} = \tfrac{s}{s^2+b^2}.

Tip
Derivatives become multiplication

L{y′}=s L{y}−y(0)\mathcal L\{y'\} = s\,\mathcal L\{y\} - y(0): differentiation turns into multiplying by ss, with the initial condition built in automatically. An initial-value problem becomes ALGEBRA in ss — solve for L{y}\mathcal L\{y\}, split by partial fractions, read the table backward.

Example
Reading the table backward

F(s)=6s4F(s) = \tfrac{6}{s^4}: since L{t3}=3!s4=6s4\mathcal L\{t^3\} = \tfrac{3!}{s^4} = \tfrac{6}{s^4}, the inverse transform is f(t)=t3f(t) = t^3. Recognizing table entries inside partial-fraction pieces is the entire skill.

Side note
Significance: the engineer’s transform

Control theory runs on Laplace: a system's transfer function H(s)H(s) IS the transform of its impulse response, and stability is "poles in the left half-plane" — statements about ss, not tt. Circuit analysis, signal processing, and PID controller design all speak this language natively.

Try it
Try it: one transform, no tables open

Compute L{5e−2t+3t}\mathcal L\{5e^{-2t} + 3t\}. Work: linearity: 5s+2+3s2\tfrac{5}{s+2} + \tfrac{3}{s^2}. Inverse practice: what has transform 4s2+4\tfrac{4}{s^2+4}? Work: bs2+b2\tfrac{b}{s^2+b^2} with b=2b = 2, scaled by 22: 2sin⁡2t2\sin 2t.

Proofs & Why It Matters

Tip
Proof: Le^at = 1/(s−a)

Directly: ∫0∞eate−stdt=∫0∞e−(s−a)tdt=[e−(s−a)t−(s−a)]0∞=1s−a\int_0^\infty e^{at}e^{-st}dt = \int_0^\infty e^{-(s-a)t}dt = \left[\dfrac{e^{-(s-a)t}}{-(s-a)}\right]_0^\infty = \dfrac{1}{s-a} for s>as > a (the boundary term at ∞\infty dies because s−a>0s - a > 0). Setting a=0a = 0 gives L{1}=1s\mathcal L\{1\} = \tfrac1s as a corollary. ■\blacksquare

Tip
Proof: Ly′ = sLy − y(0)

Integrate by parts: ∫0∞y′(t)e−stdt=[y(t)e−st]0∞+s∫0∞y(t)e−stdt=−y(0)+s L{y}\int_0^\infty y'(t)e^{-st}dt = \left[y(t)e^{-st}\right]_0^\infty + s\int_0^\infty y(t)e^{-st}dt = -y(0) + s\,\mathcal L\{y\}, assuming yy grows slower than este^{st} so the boundary term at infinity vanishes.

Iterating gives L{y′′}=s2L{y}−sy(0)−y′(0)\mathcal L\{y''\} = s^2\mathcal L\{y\} - sy(0) - y'(0) — initial conditions enter automatically, which is the method's superpower. ■\blacksquare

Tip
Proof: Ltⁿ = n!/s^n+1

Induction with parts: L{tn}=∫0∞tne−stdt=ns∫0∞tn−1e−stdt=ns L{tn−1}\mathcal L\{t^n\} = \int_0^\infty t^ne^{-st}dt = \tfrac ns\int_0^\infty t^{n-1}e^{-st}dt = \tfrac ns\,\mathcal L\{t^{n-1}\} (boundary terms vanish). Unwinding from L{1}=1s\mathcal L\{1\} = \tfrac1s: L{tn}=ns⋅n−1s⋯1s⋅1s=n!sn+1\mathcal L\{t^n\} = \tfrac{n}{s}\cdot\tfrac{n-1}{s}\cdots\tfrac{1}{s}\cdot\tfrac1s = \tfrac{n!}{s^{n+1}}. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
The transform pipeline, and where each rule enters

Solving an IVP by Laplace is a fixed pipeline. (1) TRANSFORM the equation term by term: L{y′}=sY−y(0)\mathcal L\{y'\} = sY - y(0) and L{y′′}=s2Y−sy(0)−y′(0)\mathcal L\{y''\} = s^2Y - sy(0) - y'(0) fold the initial conditions in at the start — no constants to find later.

(2) SOLVE the resulting ALGEBRAIC equation for Y(s)Y(s) — always possible, it is linear in YY. (3) PARTIAL-FRACTION YY into table-sized pieces: 1(s−a)(s−b)=1a−b(1s−a−1s−b)\tfrac{1}{(s-a)(s-b)} = \tfrac{1}{a-b}\left(\tfrac{1}{s-a} - \tfrac{1}{s-b}\right), quadratic pieces aimed at the sine/cosine entries. (4) INVERT with the table. The bookkeeping wins where classical methods strain: piecewise forcing (step functions), impulses, and long chains of initial conditions all ride through the algebra unchanged. The cost is fluency with a small table — and partial fractions, which do the heavy lifting in every problem.

Example
Worked: an IVP through the full pipeline

Solve y′+y=1y' + y = 1, y(0)=0y(0) = 0.

Step 1 — transform both sides: (sY−y(0))+Y=1s(sY - y(0)) + Y = \tfrac1s, so (s+1)Y=1s(s + 1)Y = \tfrac1s.

Step 2 — algebra: Y=1s(s+1)Y = \dfrac{1}{s(s+1)}.

Step 3 — partial fractions: write 1s(s+1)=As+Bs+1\dfrac{1}{s(s+1)} = \dfrac{A}{s} + \dfrac{B}{s+1}; multiplying out, 1=A(s+1)+Bs1 = A(s+1) + Bs; set s=0s = 0: A=1A = 1; set s=−1s = -1: B=−1B = -1.

So Y=1s−1s+1Y = \tfrac1s - \tfrac{1}{s+1}.

Step 4 — invert with the table: y(t)=1−e−ty(t) = 1 - e^{-t}.

Step 5 — verify: y′=e−ty' = e^{-t} and y′+y=e−t+1−e−t=1y' + y = e^{-t} + 1 - e^{-t} = 1 ✓, y(0)=0y(0) = 0 ✓. Compare with the integrating-factor route: same answer, but here the initial condition entered at Step 1 and no constant CC ever appeared.

Example
Worked: a second-order IVP with a sine, by table and partial fractions

Solve y′′+y=0y'' + y = 0, y(0)=0y(0) = 0, y′(0)=3y'(0) = 3 — deliberately simple, to watch the machinery.

Step 1 — transform: s2Y−s⋅0−3+Y=0s^2Y - s\cdot0 - 3 + Y = 0.

Step 2 — solve: Y=3s2+1Y = \dfrac{3}{s^2 + 1}.

Step 3 — no partial fractions needed: this is 33 times the table entry L{sin⁡t}=1s2+1\mathcal L\{\sin t\} = \tfrac{1}{s^2+1}.

Step 4 — invert: y=3sin⁡ty = 3\sin t.

Step 5 — verify: y′′=−3sin⁡t=−yy'' = -3\sin t = -y ✓, y(0)=0y(0) = 0 ✓, y′(0)=3cos⁡0=3y'(0) = 3\cos 0 = 3 ✓. Now the same skeleton with forcing, damping, or a step function only lengthens

Step 3 — the pipeline itself never changes shape, which is exactly its virtue.