Auxiliary equations with repeated roots

Second order differential equations, characteristic equation has repeated roots, reduction of orders,
blackpenredpen

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[@earl8295]

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[@l3igl2eaper]

Will going through the reduction of orders process always yield all the solutions?

[@phenylalanine8145]

Thank you that was a good explaination but what about if the root was negative ?

[@joluju2375]

I'd like to know, too.

[@krs-fltutorials4487]

@@joluju2375 Complex numbers.

[@benwinstanleymusic]

Khan academy has a video addressing that, it involves Euler's formula

[@carultch]

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.

[@phenylalanine8145]

@@carultch thank you for your answer

[@Seeker-pq4ec]

Why integration of zero is constant?

[@aljoker3053]

A coil of inductance L Henry and a capacitor of C Farad are connected in series if I=I0, Q=Q0 when t=0 Find: Q and I when t

[@carultch]

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))

[@Playboyy1985]

so do you always just add t to e^t or x to e^x ?

[@carultch]

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)

[@GHOSTRIDER4EVER]

That screen change tho: yeeeeet he knows me now and my mind too

[@danrobertluarez4318]

Teacher When your put the value of the double prime to the orginal formula why dont you square the first term like the original formula?

[@ioannaathineos5269]

KING

[@joanchepkorir6787]

What if we have been given conditions

[@urano4810]

Are auxiliary equations the same thing as homogeneous equations?

[@soundofthew4v3s]

yes

[@teotyrov]

how do you know that there are no other solutions?

[@AndDiracisHisProphet]

nth order differential equations have n different (i.e. linearly independent) solutions. since he found two, you know there are no other solutions

[@joluju2375]

@@AndDiracisHisProphet Thanks, I didn't know that. When were you taught that, is there a video demo somewhere ?

[@AndDiracisHisProphet]

@@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.

[@joluju2375]

​@@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.

[@AndDiracisHisProphet]

@@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

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HI How can v'' e^3t never becomes zero? @blackpenredpen

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e to any power is never zero, try to find x so that e^x equals 0, you won't find it, it doesn't exist

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and this is because e is a constant So e^1=e And e^0=1 Friendly reminder

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