5.2 Traditional Differential Equations Approach
Substituting i(t) = ICcos3t + ISsin3t into the differential equation 2(di(t)/dt) + 3i(t) = 10cos(3t) yields:
2(-3ICsin3t + 3IScos3t) + 3(ICcos3t + ISsin3t) = 10cos(3t) or …
(3IC + 6IS)·cos3t + (-6IC + 3IS)·sin3t = 10·cos3t + 0·sin3t
Equating coefficients yields:
3IC + 6IS = 10 and -6IC + 3IS= 0
Which have solution IC » 0.667 and IS » 1.33
So a particular solution is i(t) » 0.667·cos3t + 1.33·sin3t
However, A·cos(wt) + B·sin(wt) = (A2 + B2)1/2 cos(wt - tan-1(B/A))
\ i(t) » 1.491cos(3t - 63.43°)