Can you spot the error?

I was stuck on this problem for not 1, not 2 but 3 nights!

... Can you spot the stupid error I made?

Problem:
Let w = 2000
V = 6VL = 0.2H
R = 400

Given the differential equation L*(di/dt) + R*i = V*cos(w*t)

Find i(t). - The above equation is true for t > 0

NOTE: When im integrating ive used the '!' symbol

For the transient part:

-> 0.2*(di/dt) + 400*i = 6*cos(w*t)
-> (di/dt) + 2000*i = 30*cos(w*t)
-> di/dt + 2000*i = 0
-> di = -2000*i*dt
-> !(1/i)*di = !-2000*dt
-> ln(i) = -2000t + c
-> i(transient) = e^(-2000t)*e^c (Let e^c = A (a constant))
-> i(transient) = A*e^-2000t

For the steady state part:

Ohms law states that the current i = V/Z, where V is voltage, and Z is impedance.
So, V = 6>0 V. The '>' is the polar form (or phasor some might know it as) representation of the complex angle associated with the voltage, so 6V, 0 angle.

Z = R + j*w*L
Z = 400 + j*2000*0.2
Z = 400 + j400Z = 565.685>0.785 (all angles are in radians)

-> i = V/Z-> i = 6>0 / 565.685>0.785
-> i = 10.61>-0.785 mA
-> i(steady state) = 10.61*sin(w*t - 0.785) mA

To derrive i(t):

-> i(t) = i(steady state) + i(transient)
-> i(t) = A*e^-2000t + 10.61*sin(w*t - 0.785) mA

To find A make t = 0

-> i(0) = A(1) + (10.61*10^-3)*sin(-0.785)
-> 0 = A - 7.5*10^-3
-> A = 7.5*10^-3

So the final equation for i(t) is:
-7.5*e^-2000t + 10.61*sin(w*t - 0.785) mA

Trialling this solution by substitution does not satisfy the equation!