L75 - Fall 2003 Math 308/501502 7 Laplace Transforms 7.5...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Fall 2003 Math 308/501–502 7 Laplace Transforms 7.5 Solving Initial Value Problems Wed, 15/Oct c ± 2003, Art Belmonte Summary We now have all the pieces in place to solve initial value problems via Laplace transforms. There are four major steps involved (whether hand or machine techniques are used). If you are doing problems by hand, there may be several substeps involved in a given step. This will be illustrated in some of the hand examples. Overview of the method 1. Take the Laplace transform of each side of the DE. 2. Substitute for the initial condition(s). 3. Solve for Y ( s ) , the Laplace transform of y ( t ) . 4. Take the inverse Laplace transform of Y ( s ) to obtain y ( t ) . Hand Examples We’ll start out with a very easy problem. Example A1 Use Laplace transforms to solve the homogeneous IVP y 00 - 3 y 0 - 4 y = 0 , y ( 0 ) = 1 , y 0 ( 0 ) =- 1 Solution Here are the four steps mentioned in the overview of the method. Recall that Y ( s ) = L { y ( t ) } . 1. Now s 2 Y ( s ) - sy ( 0 ) - y 0 ( 0 ) - 3 ( sY ( s ) - y ( 0 )) - 4 Y ( s ) = 0. 2. Next, s 2 Y ( s ) - s + 1 - 3 ( ( s ) - 1 ) - 4 Y ( s ) = 0. 3. Thus ( s 2 - 3 s - 4 ) Y ( s ) = s - 4. Accordingly, Y ( s ) = s - 4 s 2 - 3 s - 4 = s - 4 ( s + 1 )( s - 4 ) = 1 s + 1 . 4. Therefore, y ( t ) = e - t . Example A2 For comparison, solve the same problem using Chapter 4 techniques. The amount of work involved is roughly the same in this instance. Solution Now r 2 - 3 r - 4 = ( r + 1 )( r - 4 ) = 0, whence r 1 , 4. Thus y = c 1 e - t + c 2 e 4 t and y 0 c 1 e - t + 4 c 2 e 4 t . The ICs give 1 = y ( 0 ) = c 1 + c 2 and - 1 = y 0 ( 0 ) c 1 + 4 c 2 Thus 0 = 5 c 2 , whence c 2 = 0and c 1 = 1. Therefore, y = e - t . Example B Use Laplace transforms to solve the initial value problem y 0 - 2 y = e - t cos t , y ( 0 ) 2 Solution Let’s recount the steps in this more complicated example. 1. Take the Laplace transform of each side of the DE. ( s ) - y ( 0 ) - 2 Y ( s ) = ( s + 1 ) ( s + 1 ) 2 + 1 2 2. Substitute for the initial condition(s). ( s ) + 2 - 2 Y ( s ) = s + 1 s 2 + 2 s + 2 3. Solve for Y ( s ) , the Laplace transform of y ( t ) . ( s - 2 ) Y ( s ) = s + 1 s 2 + 2 s + 2 - 2 Y ( s ) = - 2 s 2 - 3 s - 3 ( s - 2 )( s 2 + 2 s + 2 ) 4. Take the inverse Laplace transform of Y ( s ) to obtain y ( t ) . (a) And the sergeant gives the order that sends shivers down your spine: “Fix bayonets!” Yes, troops; it’s time for some hand-to-hand combat. We must do a partial fraction decomposition of Y ( s ) .
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/29/2008 for the course MATH 308 taught by Professor Comech during the Spring '08 term at Texas A&M.

Page1 / 4

L75 - Fall 2003 Math 308/501502 7 Laplace Transforms 7.5...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online