201.ev1.12.Lec14

# Taking l of the equation we have l a y b y c y f s 20

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Taking L of the equation we have L{ a y ′′ + b y + c y } = F ( s ) . (20) Now use linearity a L{ y ′′ } + b L{ y } + c L{ y } = F ( s ) . (21) Obviously the idea is to find L{ y } . But if we cannot represent L{ y } , L{ y ′′ } using L{ y } , we cannot proceed. Laplace transform of derivatives. Let’s start from first derivative. Example 2. Compute that Laplace transform of y . Solution. By definition L{ y } ( s ) = integraldisplay 0 e st y ( t )d t = integraldisplay 0 e st d y ( t ) = e st y ( t ) 0 integraldisplay 0 y ( t )d e st = y (0)+ s integraldisplay 0 e st y ( t )d t = y (0)+ s L{ y } ( s ) . (22) Note that in the above we have assumed lim t ր∞ [ e st y ( t )]=0 . (23) Now the general case: Theorem 3. Let y ( t ) , y ( t ) , , y ( n 1) ( t ) be continuous on [0 , ) and let f ( n ) ( t ) be piecewise continuous on [0 , ) , with all these functions of exponential order α . Then for s > α , L braceleftbig y ( n ) bracerightbig ( s )= s n L ( y )( s ) s n 1 y (0) s n 2 y (0) y ( n 1) (0) . (24) 2 Math 201 Lecture 14: Using Laplace Transform to Solve Equations

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How to solve initial value problem (IVP) using Laplace transform. From the above we see that to calculate transforms of derivatives, values of y and its derivatives at 0 are needed. Thus it is natural to consider initial value problems. Procedure: Given a y ′′ + b y + c y = f ( t ) , (25) with initial values y (0) , y (0) . Transform the equation. 1. Transform the right hand side: F ( s )= L{ f } ( s ) . (26) 2. Transform the left hand side: Denote Y ( s )= L{ y } ( s ) . L{ y ′′ } = s 2 Y s y (0) y (0); L{ y } = s Y y (0) (27) and substitute into the equations. The transformed equation is then ( a s 2 + b s + c ) Y = F ( s )+ a s y (0)+ a y (0)+ b y (0) . (28) Solve for Y ( s ) . Take inverse transform of Y ( s ) to get y . That is find y ( t ) such that L{ y } = Y . The only thing we haven’t discussed so far is the last step. Remark 4. There is indeed an inverse transform formula: L 1 { Y ( s ) } = 1 2 π i lim T →∞ integraldisplay α iT α + iT e st Y ( s )d s (29) Here α is any real number satisfying: α + β i belonds to the domain of Y ( s ) for all β. (30) As we can see that integration takes place in the complex plane C . This formula is universally
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