6.1 Laplace Transforms

Steps for solving a diffeq with initial value using Laplace transforms

1. Replace all terms in diffeq with Laplace transforms, until the only Laplace term is ℒ[y] 2. Isolate ℒ[y] on the left as its own term, and everything else on the right. 3. Use inverse Laplace transforms (and partial fractions, possibly) to find solution y (already including initial value)

Linearity of the Laplace transform

Given functions f and g and a constant c ℒ[f+g]=ℒ[f]+ℒ[g] ℒ[cf]=cℒ[f]

Laplace transform function Y of y

Y(s)=∫(0 to ∞) y(t)*e^(-st)dt for all numbers s for which this improper integral converges, and it depends on what y(t) is

Partial fractions for c/[(s+a)(s²+b)], where s²+b cannot be further factored

c/[(s+a)(s²+b)]=α/(s+a)+(βs+δ)/(s²+b)

Solving differential equations using Laplace transforms

dy/dt=y-4e^(-t), y(0)=1 is our example ℒ[dy/dt]=ℒ[y-4e^(-t)] sℒ[y]-y(0)=ℒ[y]-4ℒ[e^(-t)] Isolate ℒ[y], so that: ℒ[y]=1/(s-1)-4/[(s-1)(s+1)] Then use inverse laplace transform so that: y=ℒ⁻¹[1/(s-1)-4/[(s-1)(s+1)]] Use partial fractions to find the correct separation for the second term. This gives: y=-e^(t)+2e^(-t), and so we see that the initial value is already accounted for as well.

ℒ[1]

e⁰=1 so 1/(s-0)=1/s for s>0

Laplace transforms of derivatives

ℒ[dy/dt]=sℒ[y]-y(0)

Laplace transform of e^(at)

ℒ[e^(at)]=∫(0 to ∞)e^(at)*e^(-st)dt=∫(0 to ∞)e^[(a-s)t]dt Therefore, ℒ[e^(at)]=1/(s-a) if s>a

ℒ[t^n]

ℒ[t^(n-1)]*n/s We can derive this equation using integration by parts with u as t^n and dv/dt as e^(-st): ∫(0 to ∞)u*dv/dt=uv(0 to ∞)-∫(0 to ∞)du/dt*v The bounds for uv make it come out to 0, so just the integral helps us determine the equation (it is in the form of a laplace transform of t^(n-1)

Inverse Laplace transform

ℒ⁻¹[F]=f if and only if ℒ[f]=F