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yaro99
Group Title
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
 one year ago
 one year ago
yaro99 Group Title
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
 one year ago
 one year ago

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yaro99 Group TitleBest ResponseYou've already chosen the best response.0
^Didn't know it would come out like that, you gotta copy and paste the whole link.
 one year ago

yaro99 Group TitleBest ResponseYou've already chosen the best response.0
Third line should say b_1, not b_2
 one year ago
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