5.4 Laplace Transformation
The Laplace transform, developed in the next section, turns derivatives into multiplication by \(s\): \(\mathcal L\{x'\} = sX(s) - x(0)\), where \(X = \mathcal L\{x\}\). Applied to a linear system with constant coefficients and given initial values, it produces simultaneous linear equations for the transforms.
Example 5.4.1. Solve \(\frac {dx}{dt} = 2x - 3y\), \(\frac {dy}{dt}= y - 2x\), with \(x(0) = 8\) and \(y(0) = 3\).
Solution. Let \(X = \mathcal {L}\{x\}\) and \(Y = \mathcal {L}\{y\}\). Transforming the equations \(x' - 2x + 3y = 0\) and \(y' - y + 2x = 0\), \[sX - x(0) - 2X + 3Y = 0 ,\qquad sY - y(0) - Y + 2X = 0 ,\] that is, \[(s-2)X + 3Y = 8 ,\qquad 2X + (s-1)Y = 3 .\] Multiplying the first by \(s - 1\) and the second by \(3\) and subtracting eliminates \(Y\): \[\big [(s-1)(s-2) - 6\big ] X = 8(s-1) - 9,\qquad \text {so}\qquad (s^2 - 3s -4)X = 8s - 17 .\] Hence, by partial fractions, \[X(s) = \frac {8s - 17}{(s-4)(s+1)} = \frac {5}{s + 1} + \frac {3}{s - 4},\qquad x(t) = 5e^{-t} + 3e^{4t} .\] From the first equation, \(3y = 2x - x'\), with \(x'(t) = -5e^{-t} + 12 e^{4t}\), so \[3y = 2 ( 5e^{-t} + 3e^{4t}) - (-5e^{-t} + 12e^{4t}) = 15e^{-t} - 6e^{4t} ,\qquad y(t) = 5e^{-t} - 2e^{4t} .\] As a check, \(x(0) = 8\) and \(y(0) = 3\).
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