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u = e-S dv = dfdt
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du = --sems dt v = f(t)
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* If f(t) is discontinuous at t = 0, f(0) should he evaluated at t = O , i.e., just to the right of the origin. Since we shall seldom want to differentiate functions that are discontinuous at the origin, this detail is not of great importance. However, the reader is cautioned to watch carefully for situations in which such discontinuities occur.
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THE LAPLACE TRANSFORM
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Since I we have = -f(O) + sf(s) udv = uvI vdu
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The salient feature of this transformation is that whereas the function of t was to be differentiated with respect to t, the corresponding function of s is merely multiplied by S. We shall find this feature to be extremely useful in the solution of differential equations. To find the transform of the second derivative we make use of the transform of the first derivative twice, as follows:
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L[$$ =L{-$(Z)} = sL{$f]-~I,=,
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= s[sf(s) - ml - f (O) = s2f(s) - sf(0) - f (0)
where we have abbreviated
df 0)
= f'(O)
In a similar manner, the reader can easily establish by induction that repeated application of Eq. (2.2) leads to
L d f dt I-1
= s*f(s) - s*-lf(~) _ p-*f(l)(o) _ . . . _ sf(n-*)(o) -
p-l)(o)
where f (0) indicates the ith derivative of f(t) with respect to t, evaluated for t = 0. ( ) Thus, the Laplace transform may be seen to change the operation of differentiation of the function to that of multiplication of the transform by S, the number of multiplications corresponding to the number of differentiations. In addition, some polynomial terms involving the initial values of f(t) and its first (n - 1) derivatives are involved. In later applications we shall usually define our variables so that these polynomial terms will vanish. Hence, they are of secondary concern here.
Example 2.2. Find the Laplace transform of the function n(t) that satisfies the differential equation and initial conditions
x(O) = dx (0) = d*x(O) = o dt dt*
THE LAPLACE
TRANSFORM
It is permissible mathematically to take the Laplace transforms of both sides of a differential equation and equate them, since equality of functions implies equality of their transforms. Doing this, there is obtained ,3x(s) - 2x(O) - sir (O) - x (0) + 4[s%(s) - sx(0) - x (O)] + S[sx(s) - x(O)] + 2x(s) = 5
where x(s) = L{x(r)}. Use has been made of the linearity property and of the fact that only positive values of t are of interest. Inserting the initial conditions and solving for x(s) x(s) =
s(s3 2 + 4s2 + 5s + 2) (2.3)
This is the required answer, the Laplace transform of x(t).
Solution of Differential Equations
There are two important points to note regarding this last example. In the first place, application of the transformation resulted in an equation that was solved for the unknown function by purely algebraic means. Second, and most important, if the function x(t), which has the Laplace transform 2/s(s 3 + 4s2 + 5s + 2) were known, we would have the solution to the differential equation and boundary conditions. This suggests a procedure for solving differential equations that is analogous to that of using logarithms to multiply or divide. To use logarithms, one transforms the pertinent numbers to their logarithms and then adds or subtracts, which is much easier than multiplying or dividing. The result of the addition or subtraction is the logarithm of the desired answer. The answer is found by reference to a table to find the number having this logarithm. In the Laplace transform method for solution of differential equations, the functions are converted to their transforms and the resulting equations are solved for the unknown function algebraically. This is much easier than solving a differential equation. However, at the last step the analogy to logarithms is not complete. We obviously cannot hope to construct a table containing the Laplace transform of every function f(t) that possesses a transform. Instead, we shall develop methods for reexpressing complicated transforms, such as x(s) in Example 2.2, in terms of simple transforms that can be found in Table 2.1. For example, it is easily verified that the solution to the differential equation and boundary conditions of Example 2.2 is x(t) = 1 - 2te- - e-2r The Laplace transform of x, using Eq. (2.4) and Table 2.1, is 1 - 1 -4s) = ; - zcs + 1)2 s+2 (2.4)
(2.5)
Equation (2.3) is actually the result of placing Eq. (2.5) over a common denominator. Although it is difficult to find x(t) from Eq. (2.3), Eq. (2.5) may be easily
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