Recent content by David Koufos

Ok. I tried that and got the right diagonal matrix. Thank you. I mixed up ##CMC^{T}## with ##C^{T}MC##

Ok I found the diagonal now. Thank you. I mixed up ##CMC^{T}## with ##C^{T}MC##

Ok. I tried that and got the right diagonal matrix. Thank you. I mixed up ##CMC^{T}## with ##C^{T}MC##

I just tried that with this specific example, and using my 2nd eigen vector ##(-1, 2)## I do indeed get that [M]u = λu. So now I'm just more puzzled. I don't understand why my eigen vector doesn't yield the desired diagonal matrix. I know that you can just assume the diagonalized matrix is just...

For ##M = \begin{pmatrix} 2 & 2\\ 2 & -1 \end{pmatrix}## I found the characteristic equation: ##( λ - 3 )( λ + 2) \therefore λ = 3,-2##Going back we multiply $$\begin{pmatrix} 2 - \lambda & 2\\ 2 & -1 - \lambda \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}$$ Which gives \begin{matrix} 2x -...

Would that make a big difference tho? I could always just set up the constant to be trivial.

Thank you by the way Orodruin for mentioning that it's nonlinear. I'm doing some research on "homogeneous first-order nonlinear ordinary differential equations." I guess that's the kind of equation this is.

I treated ## \int {\frac {\mathrm {d \dot{x}}} {\mathrm {\dot{x^2}}}}## the same as ##\int \frac {\mathrm{dx}} {x^2} = \int x^{-2} \mathrm{dx} = -x^{-1}##. So then I got ##- \dot x^{-1} = \frac {\mathrm {dt}}{dx} = \frac {1}{2m}\rho CA t##. Then multiplied and divided: ##\frac {\mathrm...

Hello all, I want to say thank you in advance for any and all advice on my question. My classical mechanics textbook (Marion Thornton) has been taking me through motion for a particle with retarding forces. The example it keeps giving is: m dv/dt = -kmv which can be solved for: v = v0e-kt...