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.
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)
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))
@@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.
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.
This guy is a great math teacher!
Wish I had him as one of my profs
This guy should do magic shows. He really can read my mind. 14:12
Hahahah thanks!!
@@blackpenredpenyou're living in my head rent free
You are saving my semester!!!
Ok
Finally my concepts are getting cleared .....thanks a lot for this series of differential equations
man you are just awesome....
I think I found the best teacher for me :3
@@katzuneome to!!!
Really. This channel of math is far far better
Bro just saved me ❤
When I graduate with Engineering degree I will send it to you 🤣 you saved me through so much and still with the vids thanks man .... thank you so much
you are amazing men👏👏
This video is brilliant thank you so much
Good refresher man, keep it up...you're a great Math Teacher!!!
I will attend Calc3 next semester, thanks for an intreduction just before the new semester 😀
I didn't new it is calc 3!!!
I thought it was a easy topic (I never have calc 1)
xd
you are doing great job , way of teaching is good and simple. Thanks
very nicely done, easy to understand ...thanks alot
Will going through the reduction of orders process always yield all the solutions?
Thank you so much! your videos are always helpful and well explained.
Thank you that was a good explaination but what about if the root was negative ?
I'd like to know, too.
@@joluju2375 Complex numbers.
Khan academy has a video addressing that, it involves Euler's formula
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.
@@carultch thank you for your answer
you're amazing i owe you aloooot
Math is fun with B&Rpen... Thank you!
Why integration of zero is constant?
teacher can you do more pleasssssse !!!!! I beg you
amazing..thanks
great video!
What if we have been given conditions
so do you always just add t to e^t or x to e^x ?
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)
Are auxiliary equations the same thing as homogeneous equations?
yes
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?
That screen change tho: yeeeeet he knows me now and my mind too
Thanks alot for your help !!
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
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))
Super
KING
how do you know that there are no other solutions?
nth order differential equations have n different (i.e. linearly independent) solutions. since he found two, you know there are no other solutions
@@AndDiracisHisProphet Thanks, I didn't know that. When were you taught that, is there a video demo somewhere ?
@@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
HI How can v'' e^3t never becomes zero? @blackpenredpen
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
and this is because e is a constant
So e^1=e
And e^0=1
Friendly reminder
My professor, makes no sense
@blackpenredpen, makes sense
you are my math jesus
I love you
K stands for "Konstant"
You talk too fast and your accent doesn’t help at all, slow down please!
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.