{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

nagle_differential_equations_ISM_Part46

# nagle_differential_equations_ISM_Part46 - Exercises 7.5 32...

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

Exercises 7.5 32. Applying the Laplace transform to both sides the given equation yields s 2 Y ( s ) - as - b - 5 [ sY ( s ) - a ] + 6 Y ( s ) = L - 6 te 2 t ( s ) = - 6 ( s - 2) 2 ( s 2 - 5 s + 6 ) Y ( s ) = as + b - 5 a - 6 ( s - 2) 2 = as 3 + ( b - 9 a ) s 2 + (24 a - 4 b ) s + (4 b - 20 a - 6) ( s - 2) 2 Y ( s ) = as 3 + ( b - 9 a ) s 2 + (24 a - 4 b ) s + (4 b - 20 a - 6) ( s - 2) 2 ( s 2 - 5 s + 6) = as 3 + ( b - 9 a ) s 2 + (24 a - 4 b ) s + (4 b - 20 a - 6) ( s - 2) 3 ( s - 3) = A ( s - 2) 3 + B ( s - 2) 2 + C s - 2 + D s - 3 . (For the Laplace transforms of y and y we have used equations (7.8).) Solving for A , B , C , and D , we find that A = 6 , B = 6 , C = 3 a - b + 6 , D = b - 2 a - 6 . Hence, Y ( s ) = 6 ( s - 2) 3 + 6 ( s - 2) 2 + 3 a - b + 6 s - 2 + b - 2 a - 6 s - 3 y ( t ) = 3 t 2 e 2 t + 6 te 2 t + (3 a - b + 6) e 2 t + ( b - 2 a - 6) e 3 t . 34. By Theorem 6 in Section 7.3, L t 2 y ( t ) ( s ) = ( - 1) 2 d 2 ds 2 [ L { y ( t ) } ( s )] = d 2 ds 2 [ L { y ( t ) } ( s )] . (7.9) Theorem 5 in Section 7.3 says that L { y ( t ) } ( s ) = s 2 Y ( s ) - y (0) s - y (0) . Substituting this equation into (7.9) yields L t 2 y ( t ) ( s ) = d 2 ds 2 s 2 Y ( s ) - y (0) s - y (0) = d 2 ds 2 s 2 Y ( s ) = d ds s 2 Y ( s ) + 2 sY ( s ) = s 2 Y ( s ) + 4 sY ( s ) + 2 Y ( s ) . 36. We apply the Laplace transform to the given equation and obtain L { ty } ( s ) - L { ty } ( s ) + L { y } ( s ) = L { 2 } ( s ) = 2 s . (7.10) 221

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

View Full Document
Chapter 7 Using Theorem 5 in Section 7.3 and the initial conditions, we express L { y } and L { y } in terms of Y . L { y } ( s ) = sY ( s ) - y (0) = sY ( s ) - 2 , L { y } ( s ) = s 2 Y ( s ) - sy (0) - y (0) = s 2 Y ( s ) - 2 s + 1 . We now involve Theorem 6 in Section 7.3 to get L { ty } ( s ) = - d ds [ L { y } ( s )] = - d ds [ sY ( s ) - 2] = - sY ( s ) - Y ( s ) , (7.11) L { ty } ( s ) = - d ds [ L { y } ( s )] = - d ds s 2 Y ( s ) - 2 s + 1 = - s 2 Y ( s ) - 2 sY
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