MATH30650 Lecture 9: Diff eq lecture 9
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Laplace transforms f(t) --> f(s) (cid:2998) (cid:1832)((cid:1871)) = (cid:3505) (cid:1857)(cid:2879)(cid:3046)(cid:3047)(cid:1858)((cid:1872))(cid:1856)(cid:1872) (cid:2868) f(t) = 1, f(s) = 1/s f(t) = t (cid:1832)((cid:1871)) = (cid:3505) (cid:1857)(cid:2879)(cid:3046)(cid:3047) (cid:1872) (cid:1856)(cid:1872) Integration by parts: (cid:2998) (cid:2868) f(t) = t, f(s) = 1/s2. Laplace transform changes the complicated operation of differentiation into something simple (algebraic operations of multiplying by s and adding a constant) {(cid:1858)}((cid:1871)) = (cid:3505) (cid:1857)(cid:2879)(cid:3046)(cid:3047)(cid:1858) ((cid:1872))(cid:1856)(cid:1872) (cid:1856)(cid:1873) = (cid:1871)(cid:1857)(cid:2879)(cid:3046)(cid:3047) (cid:1856)(cid:1872),(cid:1874) = (cid:1858)((cid:1872)) (cid:2998) (cid:2868) Differential equations page 2 (cid:1877)(cid:4593)(cid:4593) +4(cid:1877) = 0 (cid:1877)(0) = 1 (cid:1877)(cid:4593)(0) = 1. {(cid:1877)(cid:4593)(cid:4593)}+4 {(cid:1877)} = {0} (cid:1871)(cid:2870)(cid:1851)((cid:1871)) (cid:1871)(cid:1877)(0) (cid:1877)(cid:4593)(0)+4(cid:1851)((cid:1871)) = 0 (cid:1871)(cid:2870)(cid:1851)((cid:1871)) (cid:1871) 1+4(cid:1851)((cid:1871)) = 0 (cid:1851)((cid:1871)) = (cid:1871) +1 (cid:1755)(cid:1755)(cid:1755)(cid:1755)(cid:1755)(cid:1755) (cid:1871)(cid:2870) +4. We need to plug back into the laplace transform equation. Laplace both sides of diff eq: solve (algebraically!) for y(s) (easy!!!, invert to get y(t) from y(s) (the hard part :( ) Theorem: there can be at most one f(t) that produces a given f(s)
