{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Quiz-1-Math-20D-W09-Solutions

# Quiz-1-Math-20D-W09-Solutions - y = 2-e x 3 2 y y(0 = 0 and...

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

Quiz #1 Math 20D January 16 SOLUTIONS 1. ( § 2.1, #30. 10 points) Find the value of y 0 for which the solution of the initial value problem y 0 - y = 1 + 3 sin t, y (0) = y 0 remains finite as t → ∞ . ANSWER: (1) p ( t ) = - 1 (2) μ ( t ) = e R p ( t ) dt = e - t (some students will be able to figure this out directly, that is OK) (3) Thus y ( t ) = 1 μ ( t ) Z g ( t ) μ ( t ) dt + C = e t Z e - t (1 + 3 sin t ) dt + C = e t - e - t + 3 Z e - t sin tdt + C (Give say 7 points if students get to here correctly.) (4) Compute R e - t sin tdt by integration by parts twice. In particular: Let u = sin t, dv = e - t dt, du = cos tdt v = - e - t . Then Z e - t sin tdt = - e - t sin t + Z e - t cos tdt. Now let u = cos t, dv = e - t dt, du = - sin tdt v = - e - t . Then Z e - t cos tdt = - e - t cos t - Z e - t sin tdt. Hence Z e - t sin tdt = - e - t sin t - e - t cos t - Z e - t sin tdt, so that Z e - t sin tdt = - 1 2 e - t (sin t + cos t ) . So y ( t ) = e t - e - t + 3 Z e - t sin tdt + C = - 1 - 3 2 (sin t + cos t ) + Ce t . Now y (0) = y 0 , so that y 0 = y (0) = - 1 - 3 2 + C = - 5 2 + C. Therefore y ( t ) = - 1 - 3 2 (sin t + cos t ) + y 0 + 5 2 e t .

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

View Full Document
The only way this can be bound is if y 0 + 5 2 = 0 , that is y 0 = - 5 2 . 2. ( § 2.2, #24. 10 points) Solve the initial value problem
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: y = 2-e x 3 + 2 y , y (0) = 0 and determine where the solution attains its maximum value. ANSWER: This is a separable equation: (3 + 2 y ) dy = (2-e x ) dx, so (integrate) 3 y + y 2 = 2 x-e x + C. Since y (0) = 0 , we have 0 =-1 + C, so that C = 1 . We get 2 x-e x + 1 = 3 y + y 2 = ± y + 3 2 ¶ 2-9 4 , so that ± y + 3 2 ¶ 2 = 2 x-e x + 13 4 . We take the positive square root to get: y =-3 2 + r 2 x-e x + 13 4 . The x value where y attains its maximum is where 2 x-e x + 13 4 attains its maximum. Settin the derivative to be zero, we get 0 = d dx ± 2 x-e x + 13 4 ¶ = 2-e x . So it is where e x = 2 , that is, x = ln2 ....
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