# c28 - 18.03 Class 28 Apr 21 Laplace Transform III Second...

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18.03 Class 28, Apr 21 Laplace Transform III: Second order equations; completing the square. Rules: L is linear: L[af(t) + bg(t)] = aF(s) + bG(s) F(s) essentially determines f(t) s-shift: L[e^{at}f(t)] = F(s-a) t-shift: L[f_a(t)] = e^{-as} F(s) s-derivative: L[tf(t)] = - F'(s) t-derivative: L[f'(t)] = s F(s) - f(0+) L[f"(t)] = s^2 F(s) - s f(0+) - f'(0+) (ignoring singularities at t = 0 ) Computations: L[1] = 1/s L[e^{as}] = 1/(s-a) L[cos(omega t)] = s/(s^2+omega^2) L[sin(omega t)] L[delta(t-a)] = = omega/(s^2+omega^2) e^{-as} L[t^n] = n!/s^{n+1} , n = 1, 2, 3, ... [1] To handle second degree equations we'll need to know the LT of f"(t). We'll compute it by regarding f"(t) as the derivative of f'(t). We'll employ a technique here that will get repeated several more times today: pick a new function symbol and use it to name some function that arises in the middle of a calculation. The application of this principle here is to write g(t) = f'(t) . We'll assume that f(t) and f'(t) are continuous for t > 0 and ignore singularities at t = 0 , so that G(s) = L[g(t)] = s F(s) - f(0+) Write down the t-derivative rule for g(t): L[g'(t)] = s G(s) - g(0+) So then

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L[f"(t)] = s (s F(s) - f(0+)) - f'(0+) = s^2 F(s) - s f(0+) - f'(0+) [2] Example: x" + 2 x' + 5 x = 5 , x(0+) = 2, x'(0+) = 3 . Note that this is EASY to solve using our old linear methods: by inspection (or undetermined coefficients, or the Key Formula) xp = 1/5 is a solution; the general solution is this plus a homogeneous solution, which you choose to satisfy the initial conditions.
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