Oct 23, 2017 ... e(2+2i)t. ) = e2t(i. 1. ) (cos(2t) + i sin(2t)) = e2t(- sin(2t) cos(2t). ) + ie2t(cos(2t) sin(
2t). ) Therefore, the general solution is. -? y (t) = c1 e2t(- sin(2t) cos(2t). ) ... y sin(
x) + x2ey - y = C. Applying the initial condition y(?) = 0, we have C = ?2. y sin(x) +
x2ey - y = ?2. Question 4. (i) C. (ii). -? y (t) -?. (0. 0. ) as t -? ?. 
