{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MATH150T8

MATH150T8 - -x(0 2 sX s-x(0 2 X s = s s 2 4 Thus X s = 1 s...

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

MATH 150 Laplace Transform Cheung Man Wai Tutorial 8 The laplace transform of a function f ( t ), which we usually denote by L { f ( t ) } or F ( s ), is defined by L { f ( t ) } = F ( s ) = 0 e - st f ( t ) dt. Suppose that the functions f, f , . . . , f n are continuous, then we have L { f ( n ) ( t ) } = s n L { f ( t ) } - s n - 1 f (0) - · · · - sf ( n - 2) (0) - f ( n - 1) (0) . Example 1 Find the solution of the initial value problem x - 2 x - 3 x = 0 x (0) = 1; x (0) = 0 . By using Laplace transform,with X ( s ) = L { x ( t ) } , 0 = L ( x - 2 x - 3 x ) = s 2 X ( s ) - sx (0) - x (0) - 2( sX ( s ) - x (0)) - 3 X ( s ) = ( s 2 - 2 s - 3) X ( s ) - x (0)( s - 2) - x (0) = ( s 2 - 2 s - 3) X ( s ) - ( s - 2) . Thus, X ( s ) = s - 2 s 2 - 2 s - 3 = s - 2 ( s - 3)( s + 1) . Let a, b R such that s - 2 ( s - 3)( s + 1) = a s - 3 + b s + 1 . Then, s - 2 = a ( s + 1) + b ( s - 3) . 1

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

View Full Document
By substituting s = 3, 1 = a (4). Thus, a = 1 4 . And putting s = - 1 gives = 3 = b ( - 4), i.e. b = 3 4 . So, we obtain, s - 2 ( s - 3)( s + 1) = 1 / 4 s - 3 + 3 / 4 s + 1 . By checking table, we have x ( t ) = 1 4 e 3 t + 3 4 e - t . Example 2 Find the solution of the initial value problem x + 2 x + 2 x = cos 2 t x (0) = 0; y (0) = 1 . The Laplace transform of the differential equation is L ( x + 2 x + 2 x ) = L (cos 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: -x (0) + 2( sX ( s )-x (0)) + 2 X ( s ) = s s 2 + 4 . Thus, X ( s ) = 1 s 2 + 2 s + 2 + s ( s 2 + 2 s + 2)( s 2 + 4) = 1 s 2 + 2 s + 2 + 1 / 10( s-2) s 2 + 2 s + 2-1 / 10( s-4) s 2 + 4 . Note that s 2 + 2 s + 2 = ( s + 1) 2 + 1. By checking that table, we have x ( t ) = e-t sin t + 1 10 e-t cos t-3 10 e-t sin t-1 10 cos 2 t + 2 10 sin 2 t. Exercise 1 Solve x + 3 x = e 2 t , x (0) =-1. Exercise 2 Solve x 00 + 4 x = cos t , x (0) = 1, x (0) = 0. Exercise 3 Solve x 00 + 4 x = e-2 t cos t , x (0) = 1, x (0) = 0. Solution: 1.-6 5 e-3 t + 1 5 e 2 t . 2. 2 3 cos 2 t + 1 3 cos t . 3. 58 65 cos 2 t + 9 65 sin 2 t + 7 65 e-2 t cos t-4 65 e-2 t sin t . 2...
View Full Document

{[ snackBarMessage ]}

What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern