- #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) = i

_{h}+ i

_{p}. 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

3. The Attempt at a Solution

My homogeneous equation is:

i

_{h}= C

_{1}e

^{λ1t}+ C

_{2}e

^{λ2t}

Where λ

_{1}= -1 and λ

_{2}= -3

My particular solution is i

_{p}= a

_{1}*sin(ωt) + a

_{2}*cos(ωt). So i'

_{p}= ωa

_{1}*cos(ωt) - ωa

_{2}*sin(ωt) and i''

_{p}= -ω

^{2}a

_{1}*sin(ωt) - ω

^{2}a

_{2}*cos(ωt).

2*sin(ωt) = i''(t) + 4i'(t) + 3i(t) = -ω

^{2}*(a

_{1}sin(ωt) + a

_{2}cos(ωt)) + 4ω(a

_{1}cos(ωt) - a

_{2}sin(ωt)) + 3(a

_{1}sin(ωt) + a

_{2}cos(ωt))

2 = a

_{1}(3-ω

^{2}) + 4ωa

_{2}

0 = a

_{2}(3-ω

^{2}) + 4ωa

_{1}

a

_{1}= (2 - 4ωa

_{2})/(3-ω

^{2})

0 = a

_{2}((3-ω

^{2})

^{2}- 16ω

^{2}) + 8ω

I could solve for a

_{2}and then for a

_{1}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.