1. 1 Week 8 3. Applications of the LT to ODEs Theorem 1: If the Laplace transforms of f(t), f ’ (t), and f ’’ (t) exist for some s, then Alternative notation:

2. 2 Consider a linear ODE with constant coefficients. It can be solved by the LT as follows: LT (step 1) ODE for y(t) Algebraic equation for Y(s) Y(s) =... y(t) =... solve (step 2) inverse LT (step 3)

3. 3 A quick review of partial fractions: Consider where P 1 and P 2 are polys. in s and the degree of P 1 is strictly smaller than that of P 2. Assume also that P 2 is factorised. Then...

4. 4 (unrepeated linear factor) (repeated linear factor) (unrepeated irreducible quadratic factor) (repeated irreducible quadratic factor)

5. 5 Example 1: (2) (1) Solution: Step 0: Observe that where and Step 1: Take the LT of (1)...

6. 6 Step 2: hence, 11 2 Step 3: The inverse LT.

7. 7 Example 2: Using partial fractions, simplify

8. 8 4. Inversion of Laplace transformation using complex integrals Theorem 2: Let F(s) be the Laplace transform of f(t). Then where γ is such that the straight line (γ – i∞, γ + i∞) is located to the right of all singular points of F(s). Question: how do we find L –1 [ F(s)] if F(s) isn’t in the Table? Answer: using the following theorem. (3) Comment: If F(s) decays as s → ∞ or grows slower than exponentially, integral (3) vanishes for all t < 0.

9. 9 Brief review of integration in complex plane: Integrals over a closed, positively oriented contour C in a complex plane can be calculated using residues. Let a function F(s) be analytic inside C except N points s = s n where it has poles (but not branch points, etc.). where res[F(s), s n ] are the residues of F(s) at s = s n. Then For example, i.e., when traversed, interior is on the left

10. 10 Example 4: Using Theorem 2, find Example 3: Using Theorem 2, find The answer: the integral in the definition of the inverse LT diverges – hence, this transform doesn’t exist.

11. 11 Comment: If an inverse transform, F(s), doesn’t decay as s → ∞, the corresponding f(t) isn’t a well-behaved function.