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Equation (3.5) may be checked by putting the right side over a common denominator, and Eq. (3.6) by substitution into the original differential equation and initial condition. Example 3.2. Solve
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~+2~-~-2x=4+e2 x(0) = 1 Taking the Laplace x (0) = 0 x (0) = -1
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transform of both sides,
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1 [s3x(s) - s* + 11 + 2[s%(s) - s] - [$X(S) - 11 - 2.X(s) = % + s-2 Solving algebraically for x(s), x(s) = s4 - 6s2 + 9s - 8 s(s - 2)(s3 + 2s2 - s - 2)
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The cubic in the denominator may be factored, and x(s) expanded in partial fractions
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x(s) =
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s4 - 6s2 + 9s - 8 A +---B +C+D+E s+1, s+2 s-1 s(s - 2)(s + I)(s + 2)(s - 1) = s s - 2 (3.7)
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To find A, multiply both sides of Eq. (3.7) by s and then set s = 0; the result is
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-8 (-2)(1)(2)(-l) = -2
The other constants are determined in the same way. The procedure and results are ~ summarized in the following table. Result
B = 1 12 D = - 712 E = v
determine
B c D E
multiply (3.7) by s-2 s+l s+2
s - l
and set s to 2 -1 -2
c = 1%
Accordingly, the solution to the problem is x(t) = -2 + +2t
+ l&-t _ ++-2t + jet
A comparison between this method and the classical method, as applied to Example 3.2, may be profitable. In the classical method for solution of differential equations we first write down the characteristic function of the homogenepus equation:
s3 + 2s2 - s - 2 = 0
INVERSION BY PARTIAL FRACTIONS
This must be factored, as was also required in the Laplace transform method, to obtain the roots -1, -2, and + 1. Thus, the complementary solution is xc(t) = Cle- + C*e-* + C3e Furthermore, by inspection of the forcing function, we know that the particular solution has the form
x,(t) = A + Be2f
The constants A and B are determined by substitution into the differential equation and, as expected, are found to be -2 and A, respectively. Then
* x ( t ) = - 2 + fie 2t + Cle- + C2em2 + Cse
and the constants Cl, C2, and Cs are determined by the three initial conditions. The Laplace transform method has systematized the evaluation of these constants, avoiding the solution of three simultaneous equations. Four points are worth noting: 1. In both methods, one must find the roots of the characteristic equation. The roots give rise to terms in the solution whose form is independent of the forcing function. These terms make up the complementary solution. 2. The forcing function gives rise to terms in the solution whose form depends
on the form of the forcing function and is independent of the left side of the equation. These terms comprise the particular solution.
3. The only interaction between these sets of terms, i.e., between the right side and left side of the differential equation, occurs in the evaluation of the constants involved. 4. The only effect of the initial conditions is in the evaluation of the constants. This is because the initial conditions affect only the numerator of x(s), as may be seen from the solution of this example. In the two examples we have discussed, the denominator of x(s) factored into real factors only. In the next example, we consider the complications that arise when the denominator of x(s) has complex factors.
Example 3.3. Solve
$+2$+2x=2 x(0) = x (0) = 0
Application of the Laplace transform yields
x(s) = 2 s(s2 + 2s + 2)
The quadratic term in the denominator may be factored by use of the quadratic formula. The roots are found to be (-1 - j) and (-1 + j). This gives the partialfraction expansion
x(s) = 2 A B c
s(s + 1 + j)(s + 1 - j) = s + (s + 1 + j) + (s + 1 - j)
(3.8)
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where A, B, and C are constants to be evaluated, so that this relation is an identity in s. The presence of complex factors does not alter the procedure at all. However, the computations may be slightly more tedious. To obtain A, multiply Eq. (3.8) by s and set s = 0:
A = (1 + j)(l - j) = l To obtain B, multiply Eq. (3.8) by (s + 1 + j) and set s = (-1 - j):
B= 2 -1-j Ep 2 (-1 - j)(-2j) 2
To obtain C, multiply Eq. (3.8) by (s + 1 - j) and set s = (-1 + j): C = Therefore, -1-j x(s) = ; + ~
C-1 + j>Gj)
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