Transient solution
y''+3y'+3.25y=3cos(t)-1.5sin(t)
I know the homogeneous solution but I got the wrong numbers for the particular solution. Here's what I tried:
http://latex.codecogs.com/gif.latex?\dpi{200}%20y_p=a_1\cos%20t+b_1\sin%20t+a_2\cos%20t+b_2\sin%20t=(a_1+a_2)\cos%20t+(b_1+b_2)\sin%20t%20\%20a_1=(F_0)_1*\frac{m(\omega_0^2-\omega_1^2)}{m^2*(\omega_0^2-\omega_1^2)^2+\omega_1^2c^2}%20\%20\%20b_2=(F_0)_1*\frac{\omega_1c}{m^2*(\omega_0^2-\omega_1^2)^2+\omega_1^2c^2}%20\%20\similarly%20\;for\;%20a_2\;and\;b_2\;using\;%20(F_0)_2\;%20and\;%20\omega_2

Question

Transient solution
y''+3y'+3.25y=3cos(t)-1.5sin(t)
I know the homogeneous solution but I got the wrong numbers for the particular solution. Here's what I tried: 
http://latex.codecogs.com/gif.latex?\dpi{200}%20y_p=a_1\cos%20t+b_1\sin%20t+a_2\cos%20t+b_2\sin%20t=(a_1+a_2)\cos%20t+(b_1+b_2)\sin%20t%20\%20a_1=(F_0)_1*\frac{m(\omega_0^2-\omega_1^2)}{m^2*(\omega_0^2-\omega_1^2)^2+\omega_1^2c^2}%20\%20\%20b_2=(F_0)_1*\frac{\omega_1c}{m^2*(\omega_0^2-\omega_1^2)^2+\omega_1^2c^2}%20\%20\similarly%20\;for\;%20a_2\;and\;b_2\;using\;%20(F_0)_2\;%20and\;%20\omega_2

anonymous · Answer

^Didn't know it would come out like that, you gotta copy and paste the whole link.

anonymous · Answer

Third line should say b_1, not b_2