Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. Put your name on each page of your paper. A short table of integrals is included on page 2: [18 points] solve the initial value problem: y = 4t3y2, y(0) = 1/4. 4t3, which in di erential form is y 2 dy = 4t3 dt and integration then gives y 1 = t4 + c, so that solving for y gives y = 1/(t4 + c). The initial condition y(0) = 1/4 means that y = 1/4 when t = 0. This gives 1/4 = 1/c so c = 4. Hence, the solution of the initial value problem is y = 4 t4 : [18 points] find the general solution of: y 3y = 5e3t + et.
