Particular solution to a second order differential equation

rive

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Hi everyone,

A teacher gave us this equation:

\(\displaystyle A*cos(\omega t)=m\frac{d^2x(t)}{dt^2}+b\frac{dx(t)}{dt}+kx(t)\)

Then he told us to change the input and give values to m, b and k. The values must real and positive and the initial conditions are these:

\(\displaystyle x(0)=x_0\neq0\)
\(\displaystyle \frac{dx(0)}{dt}=v_0\neq0\)

I set the input to \(\displaystyle 2*sin(\omega t + \pi/4)\) and m=1, b=4 and k=3

The homogeneous equation is \(\displaystyle x_h(t)=C_1e^{-t}+C_2e^{-3t}\) and I chose the particular solution to be \(\displaystyle x_p(t)=a_1sin(\omega t + \frac{\pi}{4})+a_2cos(\omega t + \frac{\pi}{4})\) so:

\(\displaystyle \frac{dx_p(t)}{dt}=\omega a_1cos(\omega t + \frac{\pi}{4})-\omega a_2sin(\omega t + \frac{\pi}{4})\)

\(\displaystyle \frac{d^2x_p(t)}{dt^2}=-\omega^2 a_1sin(\omega t + \frac{\pi}{4})-\omega^2 a_2cos(\omega t + \frac{\pi}{4})\)

Then:

\(\displaystyle 2sin(\omega t + \frac{\pi}{4})=\frac{d^2x(t)}{dt^2}+4\frac{dx(t)}{dt}+3x(t)=-\omega^2[a_1sin(\omega t + \frac{\pi}{4})+a_2cos(\omega t + \frac{\pi}{4})]+4\omega[a_1cos(\omega t + \frac{\pi}{4})-a_2sin(\omega t + \frac{\pi}{4})]+3[a_1sin(\omega t + \frac{\pi}{4})+a_2cos(\omega t + \frac{\pi}{4})]\)

\(\displaystyle 2=-\omega^2a_1-4\omega a_2+3a_1\)

\(\displaystyle 0=-\omega^2a_2+4\omega a_1+3a_1\)

\(\displaystyle a_1=\frac{2-4\omega a_2}{3-\omega^2}\)

\(\displaystyle a_2=\frac{8\omega}{-\omega^4+22\omega^2-9}\)

And this is where I'm stuck, should I just give a value to omega and solve it? If I do I would get rad/seg at different powers and I won't be able to add them and get to a single value. I don't know what to do next. Please help.

Thank you very much in advance.
 
Hi everyone,

A teacher gave us this equation:

\(\displaystyle A*cos(\omega t)=m\frac{d^2x(t)}{dt^2}+b\frac{dx(t)}{dt}+kx(t)\)

Then he told us to change the input and give values to m, b and k. The values must real and positive and the initial conditions are these:

\(\displaystyle x(0)=x_0\neq0\)
\(\displaystyle \frac{dx(0)}{dt}=v_0\neq0\)

I set the input to \(\displaystyle 2*sin(\omega t + \pi/4)\) and m=1, b=4 and k=3

The homogeneous equation is \(\displaystyle x_h(t)=C_1e^{-t}+C_2e^{-3t}\) and I chose the particular solution to be \(\displaystyle x_p(t)=a_1sin(\omega t + \frac{\pi}{4})+a_2cos(\omega t + \frac{\pi}{4})\) so:

\(\displaystyle \frac{dx_p(t)}{dt}=\omega a_1cos(\omega t + \frac{\pi}{4})-\omega a_2sin(\omega t + \frac{\pi}{4})\)

\(\displaystyle \frac{d^2x_p(t)}{dt^2}=-\omega^2 a_1sin(\omega t + \frac{\pi}{4})-\omega^2 a_2cos(\omega t + \frac{\pi}{4})\)

Then:

\(\displaystyle 2sin(\omega t + \frac{\pi}{4})=\frac{d^2x(t)}{dt^2}+4\frac{dx(t)}{dt}+3x(t)=-\omega^2[a_1sin(\omega t + \frac{\pi}{4})+a_2cos(\omega t + \frac{\pi}{4})]+4\omega[a_1cos(\omega t + \frac{\pi}{4})-a_2sin(\omega t + \frac{\pi}{4})]+3[a_1sin(\omega t + \frac{\pi}{4})+a_2cos(\omega t + \frac{\pi}{4})]\)

\(\displaystyle 2=-\omega^2a_1-4\omega a_2+3a_1\)

\(\displaystyle 0=-\omega^2a_2+4\omega a_1+3a_1\)

\(\displaystyle a_1=\frac{2-4\omega a_2}{3-\omega^2}\)

\(\displaystyle a_2=\frac{8\omega}{-\omega^4+22\omega^2-9}\)

And this is where I'm stuck, should I just give a value to omega and solve it? If I do I would get rad/seg at different powers and I won't be able to add them and get to a single value. I don't know what to do next. Please help.

Thank you very much in advance.

Omega is the frequency of the "forcing function". So it will be a "given quantity".
 
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