- #1
riveay
- 10
- 0
Homework Statement
Solve: A*sin(ωt + Θ) = L*i''(t) + R*i'(t) + (1/C)*i(t). Where: A=2, L = 1, R=4, 1/C = 3 and Θ=45°.
Homework Equations
The system has to be solved by i(t) = ih + ip. I gave the values to A, L, R, 1/C and Θ. I can also give values to ω, but I've come to a doubt when solving the particular part. The teacher said that the initial conditions should not be 0. So I set Θ to 45°, that way when t=0, A*sin(ωt + 45°) = 1.4.
3. The Attempt at a Solution
My homogeneous equation is:
ih = C1eλ1t + C2eλ2t
Where λ1 = -1 and λ2 = -3
My particular solution is ip = a1*sin(ωt) + a2*cos(ωt). So i'p = ωa1*cos(ωt) - ωa2*sin(ωt) and i''p = -ω2a1*sin(ωt) - ω2a2*cos(ωt).
2*sin(ωt) = i''(t) + 4i'(t) + 3i(t) = -ω2*(a1sin(ωt) + a2cos(ωt)) + 4ω(a1cos(ωt) - a2sin(ωt)) + 3(a1sin(ωt) + a2cos(ωt))
2 = a1(3-ω2) + 4ωa2
0 = a2(3-ω2) + 4ωa1
a1 = (2 - 4ωa2)/(3-ω2)
0 = a2((3-ω2)2 - 16ω2) + 8ω
I could solve for a2 and then for a1 but I don't know that's the correct solution. I don't know what to do next, please help. Also, I would appreciate if anyone could recommend a book to read about differential equations.
Please let me know if you need more information or if I'm not making myself clear enough.
Thank you in advance.