{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

m2k_opm_lapspr

# m2k_opm_lapspr - Specic Properties of the Laplace Transform...

This preview shows pages 1–3. Sign up to view the full content.

Specific Properties of the Laplace Transform General properties of the Laplace Transform, such as linearity , i.e., ( L c 1 f 1 + c 2 f 2 ) ( s ) = c 1 ( L f 1 ) ( s ) + c 2 ( L f 2 ) ( s ), are very important but they are not specific to the transform because they are shared by many other mathematical operations. Here we will list some of the most important specific properties of the Laplace transform and show how they can be used to generate more examples of Laplace transforms of functions important in a variety of applications, including the solution of differential and integral equations. One very specific property of the transform, its behavior with respect to convolution , is so important that it is treated in a separate section. Property I: Laplace Transform of g ( x ) = e ax f ( x ). If ( L f ) ( s ) = ˆ f ( s ) is known then we can compute ( L g ) ( s ) ( L e ax f ( x )) ( s ) = ˆ f ( s - a ) . In words: the Laplace transform of e ax f ( x ) is obtained by substituting s - a for s in the formula for the Laplace transform of f . (Note that we usually omit the argument x , as in ( L f ) ( s ), but we include it, as in ( L e ax f ( x )) ( s ), if it is needed to express a particular formula.) This property is immediate from the computation ˆ g ( s ) = Z 0 e - sx e ax f ( x ) dx = Z 0 e - ( s - a ) x f ( x ) dx = ˆ f ( s - a ) . Example I(a) Since the Laplace transforms of cos bx and sin bx are s s 2 + b 2 and b s 2 + b 2 , respectively, the Laplace transforms of e ax cos bx and e ax sin bx must be s - a ( s - a ) 2 + b 2 and b ( s - a ) 2 + b 2 , respectively. Example I(b) For f ( x ) x , which has Laplace transform 1 s 2 , Prop- erty I shows that ( L e ax x ) ( s ) = 1 ( s - a ) 2 . 1

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

View Full Document
Property II: Laplace Transform of g ( x ) x n f ( x ) We differentiate ( L f ) ( s ) = R 0 e - sx f ( x ) dx with respect to s : d ds ( L f ) ( s ) = - Z 0 e - sx x f ( x ) dx which we can also write as ( L x f ( x )) ( s ) = - d ds ( L f ) ( s ) . This relationship can be used to compute the Laplace transform of x f ( x ) when the Laplace transform of f ( x ) is already known. Example II(a) ( L 1) ( s ) = 1 s ( L x · 1) ( s ) = - 1 s 2 = 1 s 2 . Example II(b) Since ( L e ax ) ( s ) = 1 s - a , we have ( L x e ax ) ( s ) = - d ds 1 s - a = 1 ( s - a ) 2 .
This is the end of the preview. Sign up to access the rest of the 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