{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mid1solns

# mid1solns - APPM 2360 Exam 1 7:00pm 8:30pm ON THE FRONT OF...

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

APPM 2360: Exam 1 7:00pm – 8:30pm, September 22, 2010. ON THE FRONT OF YOUR BLUEBOOK write: (1) your name, (2) your student ID number, (3) recitation section (4) your instructor’s name, and (5) a grading table. Text books, class notes, and calculators are NOT permitted. A one-page crib sheet is allowed. Problem 1: (20 points) Please solve the following differential equations and IVP’s using an ap- propriate method. (a) (3 points) y 0 = cos ( t ) y (b) (5 points) y 0 = sin ( t ) cos ( t )( y + 3) y(0)=1 (c) (6 points) y 0 = 3 y + e - 6 t y (0) = 3 (d) (6 points) y 0 = y t +1 + e 2 t ( t + 1) 2 Solution: (a) y 0 = cos ( t ) y dy y = cos ( t ) dt ln | y | = sin ( t ) + C y ( t ) = ke sin ( t ) (b) y 0 = sin ( t ) cos ( t )( y + 3) dy y + 3 = sin ( t ) cos ( t ) ln | y + 3 | = 1 2 sin 2 ( t ) y + 3 = ke 1 2 sin 2 ( t ) y = ke 1 2 sin 2 ( t ) - 3 y (0) = 1 1 = k - 3 k = 4 y ( t ) = 4 e 1 2 sin 2 ( t ) - 3 Alternate solution (also correct) y ( t ) = 4 e 1 2 e - 1 2 cos 2 ( t ) - 3

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

View Full Document
(c) y 0 = 3 y + e - 6 t y 0 - 3 y = e - 6 t μ = e R - 3 dt = e - 3 t ( y - 3 y ) e - 3 t = e - 6 t e - 3 t ye - 3 t = e - 9 t - 9 + C y = ke 3 t - e - 6 t 9 y (0) = 3 3 = k - 1 9 k = 28 9 y ( t ) = 28 9 e 3 t - e - 6 t 9 (d) y 0 - y t + 1 = e 2 t ( t + 1) μ = e R - ( t +1) - 1 dt = e - ln | t +1 | = ( t + 1) - 1 y 0 - y t + 1 = e 2 t ( t + 1) 2 ( t + 1) - 1 ( y 0 - y t + 1 ) = e 2 t ( t + 1) y ( t + 1) - 1 = 1 2 e 2 t ( t + 1) - Z 1 2 e 2 t dt y ( t + 1) - 1 = 1 2 e 2 t ( t + 1) - 1 4 e 2 t + C y ( t ) = 1 2 e 2 t ( t + 1) 2 - 1 4 e 2 t ( t + 1) + C ( t + 1) Problem 2: (20 points) Carbon-14 is an isotope that is often used by archaeologists to determine or verify the age of an artifact. An artifact known to date to 10000 BC has approximately 29.83%
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