Find the solution for the following initial value problem Solution The characteristic equation is: r2 + 2r + 5 = 0 r = -1 + 2i , -1 - 2i So the general solution is: y = Ae-tsin(2t) + Be-tcos(2t) y(18) = e-18 (Asin(36) + Bcos(36)) = 6 y\'(18) = e-18 ((-A-2B)sin(36) + (-B+2A)cos(36)) = -7 Therefore: A.sin(36) + B.cos(36) = 6e18 A(2cos(36) - sin(36)) + B(-2sin(36)-cos(36)) = -7e18 Solving this system of equations results in: A = -1/2 [(-2sin(36)-cos(36))*6e18 - cos(36)*(-7e18)] = (6sin(36) - 1/2 cos(36))e18 B = -1/2 [-(2cos(36) - sin(36))6e18 + sin(36)*(-7e18)] = (1/2 sin(36) + 6cos(36))e18 Therefore the solution is: y = (6sin(36) - 1/2 cos(36))e18e-tsin(2t) + (1/2 sin(36) + 6cos(36))e18e-tcos(2t).

1. Find the solution for the following initial value problem
Solution
The characteristic equation is:
r2 + 2r + 5 = 0
r = -1 + 2i , -1 - 2i
So the general solution is:
y = Ae-tsin(2t) + Be-tcos(2t)
y(18) = e-18 (Asin(36) + Bcos(36)) = 6
y'(18) = e-18 ((-A-2B)sin(36) + (-B+2A)cos(36)) = -7
Therefore:
A.sin(36) + B.cos(36) = 6e18
A(2cos(36) - sin(36)) + B(-2sin(36)-cos(36)) = -7e18
Solving this system of equations results in:
A = -1/2 [(-2sin(36)-cos(36))*6e18 - cos(36)*(-7e18)] = (6sin(36) - 1/2 cos(36))e18
B = -1/2 [-(2cos(36) - sin(36))6e18 + sin(36)*(-7e18)] = (1/2 sin(36) + 6cos(36))e18
Therefore the solution is:
y = (6sin(36) - 1/2 cos(36))e18e-tsin(2t) + (1/2 sin(36) + 6cos(36))e18e-tcos(2t)