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The Inverse Laplace Transform

26 The Inverse Laplace TransformWe now know how to find Laplace transforms of unknown functions satisfying various initial-value problems. Of course, it s not the transforms of those unknown function which are usuallyof interest. It s the functions, themselves, that are of interest. So let us turn to the general issueof finding a functiony(t)when all we know is its Laplace transformY(s). Basic NotionsOn Recovering a Function from Its TransformIn attempting to solve the differential equation in , we gotY(s)=4s 3,which, sinceY(s)=L[y(t)]|sand4s 3=L[4e3t] s,we rewrote asL[y(t)]=L[4e3t].From this, seemed reasonable to conclude thaty(t)= , what if there were another functionf(t)with the same Transform as 4e3t? Then we couldnot be sure whether the abovey(t)should be 4e3tor that other functionf(t). Fortunately,someone has managed to prove the following:Theorem (uniqueness of the transforms)Supposefandgare any two piecewise continuous functions on[0, )of exponential orderand having the same Laplace transforms,L[f]=L[g].]

532 The Inverse Laplace Transform! Example 26.5: In exercise25.1e on page 523, you found thatthe Laplacetransformof the solution to y′′ + 4y = 20e4t with y(0) = 3 and y′(0) = 12 is Y(s) = 3s2 −28 (s −4). s2 +4 The partial fraction expansion of this is Y(s) = 3s2 −28 (s − 4) s2 +4 A s − + Bs +C s2 +4 for some constants A, B and C .

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