This differential series is a god send. I was getting pissed at my professor because he was always explaining like it was the most obvious thing in the world, which by your way it always is, there's always an intuition to it. Only with your videos I can survive the periods in my college, since because of corona I will have three in this year, so less time for this class.
Use Kirchhoff's laws to add up the voltage around the loop. The "voltage drop" across the inductor and the voltage drop across the capacitor must add up to zero: L*q" + q/C = 0 where q is the charge on the capacitor. I have my reasons for quoting voltage drop for the inductor, since it isn't technically a voltage drop, but for our purposes, we can call it that and go through the same mental exercise as if it were a voltage drop. We are solving for q(t). Assume the solution is: q(t) = e^(r*t) This means: (L*r^2 + 1/C)*e^(r*t) = 0 Solve for when the first term equals zero: L*r^2 = -1/C r^2 = -1/(L*C) r = +/- j*sqrt(1/(L*C)), where j is the imaginary unit Let w = sqrt(1/(L*C)) Since we have imaginary solutions for r, this means the solution is a linear combination of sine and cosine. q(t) = A*cos(w*t) + B*sin(w*t) Its derivative is the current: I(t) = q'(t) = -A*w*sin(w*t) + B*w*cos(w*t) At t=0, q(t) = Q0, and I(t) = I0. This means: A = Q0, and B = I0/w So our solution is: q(t) = Q0*cos(t/sqrt(L*C)) + I0*sqrt(L*C) * sin(t/sqrt(L*C)) I(t) = -Q0/sqrt(L*C) * sin(t/sqrt(L*C)) + I0*cos(t/sqrt(L*C))
If the input to the square root is negative, you get complex roots. The solution ends up taking on the form of e^(c*t) * (A*cos(w*t) + B*sin(w*t)), where r's solutions are r = c +/- w*i, where i is the imaginary unit. In the event that your roots are both purely imaginary, then the exponential envelope term disappears, and you just get a linear combination of sine and cosine.
Yes. And if there is a thrice repeated root, you would square the t, or square the x, when setting up your three component functions. So you'd get A*e^(r*t) + B*t*e^(r*t) + C*t^2*e^(r*t)
@@joluju2375 Hm, I dunno. Maybe in 2nd Semester or maybe earlier. That is relativly elementary ODE stuff. Although I couldn't shake a proof out of my sleeve from the top of my head.
@@AndDiracisHisProphet The truth is I haven't opened a math book for 40 years and was never taught ODE. But bprp is so good at explaining things that now I'm learning a lot of new things, far above the maths level we have at 18 in my country, I don't know the american name for that level.
@@joluju2375 I'M not american either :) ODE is University level, although I had a little bit in my physics course in school. Have fun with mathematics :) bprp is really cool, although his stuff is more about calculating and less about proving
This is math not english class stop complaining. I understand him just fine. His technique and explain helps me refreshing and understanding more into the topic. This is a plus. When you study math you just have to focus and do lots of practice problems.
