(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:ye ^{-t}= 2te^{t}- 2e^{t}+ C,or

y = 2te ^{2t}- 2e^{2t}+ Ce^{t}.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}.