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

Info icon This preview shows pages 2–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 2

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 3
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern