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EXAMPLE 17 The following agree with the properties listed previously.
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LaplaceTransform[Exp[2t] Sin[3t], t, s] //ExpandDenominator 3 13 4 s + s2 LaplaceTransform[Exp[2t]Cos[3t], t, s] //ExpandDenominator 2 + s 13 4 s + s2 LaplaceTransform[a f[t] + b g[t], t, s] a LaplaceTransform[f[t], t, s] + b LaplaceTransform[g[t], t, s] LaplaceTransform[f'[t], t, s] f + s LaplaceTransform[f[t], t, s] LaplaceTransform[f''[t], t, s] s f + s2 LaplaceTransform[f[t], t, s] f'
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Ordinary Differential Equations
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The power of the Laplace transform is derived from the fact that there is a one-to-one correspondence between f (t) and L { f (t )}. This means that if L { f (t )} is known, then f (t) is uniquely determined. If F (s) = L { f (t )}, then f (t ) = L 1 { F (s)}. L 1 is called the inverse Laplace transform.
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InverseLaplaceTransform[F[var1], var1, var2] computes the inverse Laplace transform of the function F, with independent variable var1, and expresses it as a function of var2.
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InverseLaplaceTransform 1 , s, t s 3 3t InverseLaplaceTransform 31 , s, t s 8 1 t 3 t Cos[ 3 t] 3 Sin [ 3 t] 12
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Traditionally, one would factor the denominator, expand into partial fractions, and find the inverse transformation separately for each term. Mathematica does it all automatically.
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The next example illustrates how the Laplace transform can be used to solve a simple differential equation.
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d2y dy 3 + 2 y = t 2 , y '(0) = 1, y(0) = 2 . First we compute the dt dt 2 Laplace transform of both sides of the equation. This can be done in one step.
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EXAMPLE 19
Solve the differential equation
equation = y''[t] 3 y'[t]+ 2 y[t] t2; temp = LaplaceTransform[equation, t, s] 2 LaplaceTransform[y[t], t, s]+ s2 LaplaceTransform[y[t], t, s] 3(s LaplaceTransform[y[t], t, s] y) s y y' 2 s3 Then we solve for the Laplace transform satisfying the given initial conditions. temp2 = Solve[temp, LaplaceTransform[y[t], t, s]] /. {y' 1, y 2}
3 4 LaplaceTransform[y[t], t, s] 2 5 s + 2 s 2 s3 (2 3 s + s )
Next we extract the transform as a function of s. temp3 = temp2[[1, 1, 2]] 2 5 s3 + 2 s4 s3 (2 3 s + s2) Finally, we compute the inverse Laplace transform to get the solution of the equation. InverseLaplaceTransform[temp3, s, t] 1 (7 + 4 t 3 2t + 6 t + 2 t2 ) 4
As indicated at the beginning of this section, Laplace transforms are the ideal tool to use when dealing with discontinuous right-hand sides. In this context we shall find it convenient to introduce the Heaviside theta function and the unit step function.
UnitStep[x] returns a value of 0 if x < 0 and 1 if x 0. HeavisideTheta[x] returns a value of 0 if x < 0 and 1 if x > 0.
The unit step function, which we represent as u(t), offers a convenient way to define piecewise defined functions.
Ordinary Differential Equations
EXAMPLE 20 Plot the graph of g( x ) =
x if x < 1 for 0 x 2. x 3 if x 1
g[x_] = UnitStep[1 x]x + UnitStep[x 1]x3; Plot[g[x], {x, 0, 2}]
e cs It is easily shown that L {u(t c)} = and, if F (s) = L { f (t )} , then L {u(t c) f (t c)} = e cs F (s) . s These properties make it convenient to solve differential equations involving piecewise continuous functions.
2 y EXAMPLE 21 Solve d 2 3 dy + 2 y = g(t ) , y(0) = y' (0) = 0
dt dt Plot the solution for 0 t 2.
1 if 0 t < 1 where g(t ) = 0 if t > 1
For t 0, g(t)= UnitStep[1 t]. temp = LaplaceTransform[y''[t] 3 y'[t] + 2 y[t] UnitStep[1 t], t, s] 2 LaplaceTransform[y[t],t,s]+ s2 LaplaceTransform[y[t],t,s] 3(s LaplaceTransform[y[t],t,s]- y) s y y' 1 Cosh[s]+Sinh[s] r s
temp2=Solve[temp, LaplaceTransform[y[t], t, s]] /. {y' 0, y 0}
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