There are a lot of ways we can apply this for example

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There are a lot of ways we can apply this. For example, let y ( t ) = cos( bt ). Then Y ( s ) = s/ ( s 2 + b 2 ), as we already showed. Using this property, we get that L [ e at cos( bt )] = Y ( s - a ) = s - a ( s - a ) 2 + b 2 . Frequency Differentiation Let Y ( s ) be the Laplace transform of y ( t ). Then by definition Y ( s ) = 0 e - st y ( t ) dt. 82
If we then differential each side with respect to s , we get Y ( s ) = d ds 0 e - st y ( t ) dt = 0 d ds ( e - st ) y ( t ) dt = 0 - e - st ty ( t ) dt = - L [ ty ( t )] . This gives us the useful identity L [ ty ( t )] = - Y ( s ) . If this process is repeated n times, we get the more general identity L [ t n y ( t )] = ( - 1) n Y ( n ) ( s ) . Note that if we apply this identity with y ( t ) = 1, we get the Laplace transform of t n is L [ t n ] = ( - 1) n d n ds n 1 s = n ! s n +1 . Time Differentiation Let Y ( s ) be the Laplace transform of y ( t ). State the Laplace transform of y ( t ) in terms of Y . The Laplace transform for y ( t ) is L [ y ( t )] = 0 e - st y ( t ) dt. Now we do integration by parts with u = e - st and = dv = y ( t ) dt . Then du = - se - st dt and v = y . Integration by parts the gives us 0 e - st y ( t ) dt = e - st y ( t ) t = t =0 + 0 se - st y ( t ) dt = (0 - y (0)) + s 0 e - st y ( t ) dt = sY ( s ) - y (0) . This gives us the most useful property of Laplace transforms for solving differential equations: L [ y ( t )] = sY ( s ) - y (0) . Repeating this process a second time, would yield the result for the Laplace transform of the second derivative: L [ y ( t )] = s L [ y ( t )] - y (0) = s ( sY ( s ) - y (0)) - y (0) = s 2 Y ( s ) - sy 0 - y 0 . We can use this property to find the Laplace transform for sin( at ) since we already know the Laplace transform of y ( t ) = cos( at ): L [sin( at )] = - 1 a L d dt cos( at ) = - 1 a ( sY ( s ) - cos(0)) = - 1 a s 2 s 2 + a 2 - 1 = a s 2 + a 2 Action on a Differential Equation Consider a simple first-order non-homogeneous equation of the form ay + by = g ( t ) , y (0) = y 0 83
If we take the Laplace transform of each side, we get a L [ y ] + b L [ y ] = L [ g ( t )] which gives us a [ sY ( s ) - y 0 ] + bY ( s ) = G ( s ) Y ( s ) = G ( s ) + ay 0 as + b . Notice that we were able to solve for Y ( s ) algebraically after performing the Laplace transform. This is what makes Laplace transforms so useful: differential equations in the time domain become algebraic equations in the frequency domain. However, there still remains the problem of inverting Y ( s ) to recover the solution y ( t ). 3.1.5 Further Examples We have already seen that the Laplace transform for monomials with integer powers is L [ t n ] = n ! s n +1 . But what about non-integer powers? It turns out that there is a well defined way extend the idea of a factorial to any positive number, which is called the gamma function Γ( r ). The gamma function is defined by the integral Γ( r ) = 0 x r - 1 e - x dx. When evaluated at integers, the gamma function satisfies the relation Γ( n ) = ( n - 1)! which can be proven using successive integration by parts. Therefore we can generalize the Laplace transform for non-integer powers of t : L [ t r ] = Γ( r + 1) s r +1 . Other than integers, not many exact values for the gamma function are known. One exception is Γ(3 / 2) = (1 / 2)! = π 2 .

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