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barcode reader using c#.net T H E LAPLACE in Software
T H E LAPLACE Code128 Scanner In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Generate Code 128B In None Using Barcode generation for Software Control to generate, create Code 128A image in Software applications. TRANSFORM
USS Code 128 Reader In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. Draw Code 128 Code Set B In C#.NET Using Barcode drawer for .NET framework Control to generate, create Code 128 Code Set A image in .NET framework applications. 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 USS Code 128 Creation In Visual Studio .NET Using Barcode creator for ASP.NET Control to generate, create Code 128C image in ASP.NET applications. Code 128B Encoder In .NET Framework Using Barcode drawer for Visual Studio .NET Control to generate, create Code128 image in VS .NET applications. ~+2~~2x=4+e2 x(0) = 1 Taking the Laplace x (0) = 0 x (0) = 1 Making Code 128 In VB.NET Using Barcode creator for .NET framework Control to generate, create Code 128 image in Visual Studio .NET applications. Encoding Code 128 In None Using Barcode drawer for Software Control to generate, create Code 128B image in Software applications. transform of both sides, UPCA Maker In None Using Barcode encoder for Software Control to generate, create Universal Product Code version A image in Software applications. Create Bar Code In None Using Barcode maker for Software Control to generate, create bar code image in Software applications. 1 [s3x(s)  s* + 11 + 2[s%(s)  s]  [$X(S)  11  2.X(s) = % + s2 Solving algebraically for x(s), x(s) = s4  6s2 + 9s  8 s(s  2)(s3 + 2s2  s  2) Making EAN 128 In None Using Barcode creator for Software Control to generate, create GS1128 image in Software applications. Bar Code Creator In None Using Barcode generator for Software Control to generate, create bar code image in Software applications. The cubic in the denominator may be factored, and x(s) expanded in partial fractions
Generating Interleaved 2 Of 5 In None Using Barcode printer for Software Control to generate, create ANSI/AIM ITF 25 image in Software applications. Encode Code 128A In Java Using Barcode creation for Eclipse BIRT Control to generate, create Code 128 Code Set C image in Eclipse BIRT applications. x(s) = USS Code 128 Creation In Java Using Barcode generation for Java Control to generate, create Code 128C image in Java applications. Code 128 Printer In None Using Barcode encoder for Word Control to generate, create Code 128 Code Set C image in Microsoft Word applications. s4  6s2 + 9s  8 A +B +C+D+E s+1, s+2 s1 s(s  2)(s + I)(s + 2)(s  1) = s s  2 (3.7) UPC A Maker In None Using Barcode encoder for Microsoft Word Control to generate, create Universal Product Code version A image in Microsoft Word applications. Recognize Bar Code In Java Using Barcode Control SDK for Eclipse BIRT Control to generate, create, read, scan barcode image in Eclipse BIRT applications. To find A, multiply both sides of Eq. (3.7) by s and then set s = 0; the result is
Paint Bar Code In VS .NET Using Barcode creation for ASP.NET Control to generate, create bar code image in ASP.NET applications. UCC128 Recognizer In VB.NET Using Barcode decoder for .NET Control to read, scan read, scan image in VS .NET applications. 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 s2 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) THE LAPLACE
TRANSFORM
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 1j Ep 2 (1  j)(2j) 2 To obtain C, multiply Eq. (3.8) by (s + 1  j) and set s = (1 + j): C = Therefore, 1j x(s) = ; + ~ C1 + j>Gj)

