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Unformatted text preview: 25 Differentiation and the Laplace Transform In this chapter, we explore how the Laplace transform interacts with the basic operators of calculus: differentiation and integration. The greatest interest will be in the first identity that we will derive. This relates the transform of a derivative of a function to the transform of the original function, and will allow us to convert many initialvalue problems to easily solved algebraic equations. But there are other useful relations involving the Laplace transform and either differentiation or integration. So we’ll look at them, too. 25.1 Transforms of Derivatives The Main Identity To see how the Laplace transform can convert a differential equation to a simple algebraic equation, let us examine how the transform of a function’s derivative, L bracketleftbig f ′ ( t ) bracketrightbigvextendsingle vextendsingle s = L bracketleftbigg d f dt bracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle s = integraldisplay ∞ d f dt e − st dt = integraldisplay ∞ e − st d f dt dt , is related to the corresponding transform of the original function, F ( s ) = L [ f ( t ) ]  s = integraldisplay ∞ f ( t ) e − st dt . The last formula for L bracketleftbig f ′ ( t ) bracketrightbig clearly suggests using integration by parts, which you may recall via the equation integraldisplay b a u d v = u v vextendsingle vextendsingle b a − integraldisplay b a v du . To ensure that this integration by parts is valid, we need to assume f is continuous on [ , ∞ ) and f ′ is at least piecewise continuous on ( , ∞ ) . Assuming this, L bracketleftbig f ′ ( t ) bracketrightbigvextendsingle vextendsingle s = integraldisplay ∞ e − st bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright u d f dt dt bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright d v = e − st f ( t ) vextendsingle vextendsingle ∞ t = − integraldisplay ∞ f ( t ) bracketleftbig − se − st bracketrightbig dt 501 502 Differentiation and the Laplace Transform = lim t →∞ e − st f ( t ) − e − s · f ( ) − integraldisplay ∞ bracketleftbig − se − st bracketrightbig f ( t ) dt = lim t →∞ e − st f ( t ) − f ( ) + s integraldisplay ∞ f ( t ) e − st dt . Now, if f is of exponential order s , then lim t →∞ e − st f ( t ) = whenever s > s and F ( s ) = L [ f ( t ) ]  s = integraldisplay ∞ f ( t ) e − st dt exists for s > s . Thus, continuing the above computations for L bracketleftbig f ′ ( t ) bracketrightbig with s > s , we find that L bracketleftbig f ′ ( t ) bracketrightbigvextendsingle vextendsingle s = lim t →∞ e − st f ( t ) − f ( ) + s integraldisplay ∞ f ( t ) e − st dt = − f ( ) + s L [ f ( t ) ]  s , which is a little more conveniently written as L bracketleftbig f ′ ( t ) bracketrightbigvextendsingle vextendsingle s = s L [ f ( t ) ]  s − f ( ) (25.1a) or even as L bracketleftbig f ′ ( t ) bracketrightbigvextendsingle...
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 Summer '09
 Differential Equations, Equations, Derivative, Continuous function, dt, exponential order, exponential order s0

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