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LaplaceTransform[y[t],t,s
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1 Cosh[s]+ Sinh[s] s(2 3 s + s2)
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temp3 = temp2[[1, 1, 2]] 1 Cosh[s]+ Sinh[s] 3 s y[0] s(2 3 s + s2) f[t_] = InverseLaplaceTransform[temp3, s, t] 1 ( 1 + t )2 ( 1 + 1+t )2 HeavisideTheta[ 1 + t] 2 Plot[f[t], {t, 0, 2}]
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Observe that the solution is continuous, even though the equation involves a discontinuous function.
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Ordinary Differential Equations
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In physical and biological applications, we are often led to differential equations whose right-hand side, f(t), is a function of an impulsive nature, that is, f(t) has zero value everywhere except over a short interval of time where its value is positive. The Dirac delta function is an idealized impulse function. Although not a true function in the classical sense, its validity is justified by the theory of distributions, developed by Laurent Schwartz in the midtwentieth century. It is defined by the following pair of conditions:
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(t t0 ) = 0 if t t0
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(t t0 )dt = 1
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An immediate consequence of the definition is the result that f (t ) (t t0 )dt = f (t0 ) . It follows, there fore, that L { (t t0 )} = 0 e st (t t0 ) dt = e st , provided that t0 0. Otherwise its value is 0.
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DiracDelta[t] returns (t ), the Dirac delta function, which satisfies (t ) = 0 if t 0,
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(t )dt = 1.
EXAMPLE 22
DiracDelta[t]dt
f[t] DiracDelta[t 5] dt
f[5]
EXAMPLE 23
LaplaceTransform[DiracDelta[t a], t, s] a s HeavisideTheta[a] LaplaceTransform[DiracDelta[t 3], t, s] 3 s LaplaceTransform[DiracDelta[t + 3], t, s] 0
Since Mathematica does not know whether a is negative or non-negative, HeavisideTheta[a] is included in the Laplace transform.
Since we know the Laplace transform of the Dirac delta function, we can solve differential equations involving impulses much the same way as described in Example 19. The following example illustrates the method.
EXAMPLE 24 Find the solution of the differential equation
d2y dy 2 + y = (t 1), y(0) = y '(0) = 0. dt dt 2 equation = y''[t] 2 y'[t] + y[t] DiracDelta[t 1]; LaplaceTransform[y[t],t,s]+ s2 LaplaceTransform[y[t],t,s]
s 2(s LaplaceTransform[y[t],t,s] y[0]) s y[0] y'[0] s [
temp = LaplaceTransform[equation, t, s]
temp2 = Solve[temp, LaplaceTransform[y[t], t, s]] /. {y'[0] 0, y[0] 0}
LaplaceTransform[y[t],t,s]
s 1 2 s + s2
temp3 = temp2[[1, 1, 2]] s 1 2 s + s2 solution[t_] = InverseLaplaceTransform[temp3, s, t] 1+t ( 1 + t)HeavisideTheta[ 1 + t]
Ordinary Differential Equations
Plot[solution[t], {t, 0, 2}, PlotStyle Thickness[.01]]
2.5 2.0 1.5 1.0 0.5
Laplace transforms can be used to solve systems of differential equations. The technique is similar to that of a single equation, except that a different transform is defined for each dependent variable. The next example illustrates the method for solving a system of two first-order equations. It generalizes in a natural way to larger and higher-order systems.
dx + y = t EXAMPLE 25 Solve the system dt dy with initial condition x (0) = 1, y (0) = 1. 4 x + dt = 0 system = {x'[t] + y[t] t, 4 x[t] + y'[t] 0}; temp = LaplaceTransform[system, t, s] {s LaplaceTransform[x[t],t,s]+ LaplaceTransform[y[t],t,s] x[0] 1 , n s2 4 LaplaceTransform[x[t],t,s]+ s LaplaceTransform[y[t],t,s] y[0] 0} ] temp2 = Solve[temp, {LaplaceTransform[x[t], t, s], LaplaceTransform[y[t], t, s]}] /. {x[0] 1, y[0] 1}
1 s s2 LaplaceTransform[x[t],t,s] s( 4 + s2), 4 + 4 s2 + s3 LaplaceTransform[y[t],t,s] 2 s ( 4 + s2)
temp3a = temp2[[1, 1, 2]] 1 s s2 s( 4 + s2) temp3b = temp2[[1, 2, 2]]
2 3 + 42 4 s + s s ( 4 + s2)
InverseLaplaceTransform[temp3a, s, t] 2t 7 2t 1 + 3 + 4 8 8 InverseLaplaceTransform[temp3b, s, t]
2 t 2t 3 7 + t 4 4 The solution to the system is 2 t 2t 2 t 2t x = 1 + 3e + 7 e , y = 3 e 7 e + t 4 8 8 4 4
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