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barcode reader using c#.net Proof. in Software
Proof. Read USS Code 128 In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Encoding Code128 In None Using Barcode encoder for Software Control to generate, create Code 128 Code Set B image in Software applications. To integrate this by parts, let
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Code128 Generator In VB.NET Using Barcode creator for .NET Control to generate, create Code128 image in Visual Studio .NET applications. Encode Bar Code In None Using Barcode generation for Software Control to generate, create barcode image in Software applications. du = sems dt v = f(t) Printing Code128 In None Using Barcode generator for Software Control to generate, create Code 128 Code Set A image in Software applications. Bar Code Creator In None Using Barcode drawer for Software Control to generate, create bar code image in Software applications. * 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. Painting DataMatrix In None Using Barcode maker for Software Control to generate, create Data Matrix 2d barcode image in Software applications. Encode GS1128 In None Using Barcode encoder for Software Control to generate, create GTIN  128 image in Software applications. THE LAPLACE TRANSFORM
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Printing 1D In Java Using Barcode creation for Java Control to generate, create Linear Barcode image in Java applications. Creating Barcode In None Using Barcode creation for Online Control to generate, create barcode image in Online applications. 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: Code39 Encoder In Visual Basic .NET Using Barcode maker for VS .NET Control to generate, create Code 39 image in .NET applications. ECC200 Recognizer In C# Using Barcode reader for .NET framework Control to read, scan read, scan image in Visual Studio .NET applications. L[$$ =L{$(Z)} = sL{$f]~I,=, GTIN  13 Drawer In .NET Using Barcode creator for ASP.NET Control to generate, create EAN 13 image in ASP.NET applications. EAN 13 Generator In Java Using Barcode drawer for Android Control to generate, create GTIN  13 image in Android applications. = 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 I1 = s*f(s)  s*lf(~) _ p*f(l)(o) _ . . . _ sf(n*)(o)  pl)(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  e2r 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

