This is an augmented matrix. It’s a shorthand way of writing a system of equations that has non zeros on the right side in matrix form. You ought to have covered this in a linear algebra course before taking an ODE course but if not just write down the two equations and solve them simultaneously.
Do you mean if you had a 2nd eigenvector (so the matrix isnt defective) It'd be x(t) = c1e^2t * v1 + c2e^2t * v2 for the general solution? (v1,v2 are the 2 eigenvectors if it wasnt defective) Or do we: c1e^2t * v1 + tc2*e^2t * v2 (Multiply a t in front of 2nd term)?
Awesome thanks! I'm in a discrete course and we're working with recurrence relations and I had a 3x3 that wasn't defective but had one eigenvalue with a repeat root so I assume it's the same as the ODE case
I attend Stanford University and this is way better than how my TA explains it. Well done sir!
What year does Stanford typically have students learning this level of mathematics? I assume sophomore year?
Just saved me from hours of confusion doing my hw. Thank you very much sir!
Thank you very much. You cleared my confusion.
Very clear and useful!
Hello, can someone explain me what is he is doing at 10:30 with this "double matrix" thank you
This is an augmented matrix. It’s a shorthand way of writing a system of equations that has non zeros on the right side in matrix form. You ought to have covered this in a linear algebra course before taking an ODE course but if not just write down the two equations and solve them simultaneously.
Thank you so much sir. You helped me a lot.
THANKS SO MUCH VERY HELPFUL MY PROF DOESNT TEACH
save my life before exam，thank you
How do you calculate the wronskian in this kind of problem?
appreciate this video so so much
Thanks for the video, it makes much more sense to me now! :)
Great!
How you take the Value of (A-2I)={1 1 ;0 0}
Can you explain 9:03 in another way. I have no idea how you got v = (1 -1)
same
Maybe coming a bit late for you but, for others with the same question, I find the eigenvalues at 1:50 - 3:10.
Shouldn't v = [-1,1] ?
Eigenvectors aren't unique. In fact, you can always multiply an eigenvector by a constant and it will still be an eigenvector. So you're correct.
Do you mean if you had a 2nd eigenvector (so the matrix isnt defective)
It'd be x(t) = c1e^2t * v1 + c2e^2t * v2 for the general solution? (v1,v2 are the 2 eigenvectors if it wasnt defective)
Or do we:
c1e^2t * v1 + tc2*e^2t * v2 (Multiply a t in front of 2nd term)?
Yes, if the matrix had been such that there were two eigenvectors for lambda=2 then the solution would be the first x(t) that you wrote.
Awesome thanks!
I'm in a discrete course and we're working with recurrence relations and I had a 3x3 that wasn't defective but had one eigenvalue with a repeat root so I assume it's the same as the ODE case
I just noticed you have some PDE videos too...
Time to check those out! :D
thank you so much
Thank you!!
Mate I’m in oxford and my lectures on this are shit
Thanks sir
Thanks!
God
Bro get some water.