(21 pts) Solve the IVP:

y' - y = 2te^2t, y(0) = 1

Solution

This is a liner equation of the first order. There's an obvious integrating factor: e^-t. Multiply by e^-t to get: (ye^-t)' = 2te^t. Integrate both sides:

or

All that remains is to satisy the initial condition. Plug in t = 0: y(0) = -2 + C = 1. C = 3.

Answer: y = 2te^2t - 2e^2t + 3e^t.