Differential Equations Test #3 Review

• §7.2: Integral Transforms and Definition of ℒ{f(t)}
  • An "integral transform" turns f(t) (a function of t) into ∫_α^βK(s,t)·f(t)dt (a function of s)
  • The Laplace Transform specifically: ℒ{f(t)} = F(s) = ∫₀^∞e^-st·f(t)dt.
  • Evaluating this usually involves integration by parts.
  • Know the Laplace transforms of 1, t, tⁿ, e^at, sin(bt), cos(bt), e^at·sin(bt), and e^at·cos(bt).


• §7.3: Properties of the Laplace Transform
  • First derivative: ℒ{f′(t)} = s·ℒ{f(t)} − f(0)
  • Second derivative: ℒ{f″(t)} = s²·ℒ{f(t)} − s·f(0) − f′(0)
  • ℒ is linear (can split up over addition and pull out constant multipliers)
  • If ℒ{f(t)} = F(s), then ℒ{e^at·f(t)} = F(s−a)


• §7.4: Inverting the Laplace Transform
  • i.e. when you know F(s) (or ℒ{f(t)}) and you want to find f(t) (or ℒ^-1{F(s)})
  • Look for something in the s-domain column of the Laplace table that "looks like" F(s).
  • If nothing looks familiar, try breaking F(s) into partial fractions.
  • If denominator includes an irreducible quadratic, try completing the square.
    • This should make it resemble the transform of e^at·sin(bt) or e^at·cos(bt).
    • It may need to be split further into a linear numerator and a constant numerator.
  • Don't forget that ℒ is linear!


• §7.5: Solving Initial Value Problems
  • "Put it into the magic box" by taking the Laplace transform of both sides.
  • Use the expansion formulas for ℒ{y″} and ℒ{y′} (with initial conditions).
  • The differential equation is now an algebraic equation with no derivatives!
  • Solve algebraically for ℒ{y}, and decompose into partial fractions as needed.
  • "Take it back out of the magic box" by taking the inverse Laplace transform of both sides.


• §7.6: Discontinuous Functions and the unit step function u(t−a)
  • u(t) is the unit step function; u(t) = 0 if t < 0 and u(t) = 1 if t > 0.
  • Thus M·u(t−a) = 0 if t < a and M·u(t−a) = M if t > a.
  • Any piecewise-defined function can be written in terms of u(t−a)!
  • It may help to think of u as a multiplier that "turns on" a function at t=a.
  • Laplace transform for constant M: ℒ{M·u(t−a)} = M·e^-as/s
  • Non-constant: ℒ{f(t−a)·u(t−a)} = e^-as·ℒ{f(t)};   ℒ{g(t)·u(t−a)} = e^-as·ℒ{g(t+a)}


• §7.8: Impulse and the Dirac Delta Function δ(t−a)
  • impulse = F·Δt = m·Δv = Δmomentum
  • δ_τ(t−a) = 1/(2τ) on t ∈ [a−τ, a+τ]; 0 elsewhere
  • The limit of this as τ→0 is δ(t−a) = ∞ when t = a; 0 elsewhere, but total impulse is still one!
  • This is a useful approximation of a "short sharp shock" with impulse=1.
  • Laplace transform: ℒ{δ(t−a)} = e^-as
  • When solving an initial value problem, this often leads to u(t-a) in the solution.

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