APPM 2360 Midterm: appm2360summer2015exam11_sol
Document Summary
On the front of your bluebook write: (1) your name, (2) your student id number, (3) your instructor"s name, and (4) a grading table. Text books, class notes, and calculators are not permitted. Problem 1: (20 points) (a) find all solutions to (y = (3 2y) cos(t) y(0) = 1 (b) the ode 1 t z + t2z = 0 has solution: zh(t) = ce t4/4. Parameters to solve z + t2z = e t4/4. Solution 1: (a) first, we see that there is an equilibrium solution ye(t) = 3/2, which does not satisfy the initial condition. Now using separation of variables, we see that. Using our initial condition we nd that c = 1 which gives a nal answer of: dy. 2 ln(3 2y) = sin t + c. 2(cid:0)3 e 2 sin(t)(cid:1) . y(t) = 1 (b) variation of parameters gives us a guess for a solution to the non homogeneous ode based on the homogeneous solution.
