When solving the first order ODE: z' = k z where k is some number, and z is a function of t, the solution is z=ce^(kt), where c is an arbitrary constant. You can convince yourself this is true by taking the derivative of this solution and seeing that indeed z'= k z. Hope that helps!
My boy Tyler, I know its been two years since posting this, but you just helped me and my classmate out so much. Good looks bruv GG.
I never found a vid that tied together the matrix diagonalization with actually solving the Diff EQ. This is the best!!
This is the best video on solving system of simultaneous linear differential equations by using the method of diagonalization. Thanks
Thank you! You helped me out.
Tyler, you are a goat
Clear and concise explanation. Thanks so much!
The best video of Them all!
Nice video, very clear and to the point. I have one question- where does 'e' come from when you are solving the Z vectors?
When solving the first order ODE: z' = k z where k is some number, and z is a function of t, the solution is z=ce^(kt), where c is an arbitrary constant. You can convince yourself this is true by taking the derivative of this solution and seeing that indeed z'= k z. Hope that helps!
This was the best one ever.
Awesome!
So, why do you even need to solve for P inverse if you don't use it in your solution?
we need to be sure that it is has a solution so we can use it in the z = P^-1y.
Thank you.
ur a boss
how did you find limda 3 and 1?
🙏🙏🙏
Thanks
why we use [-1 1] when we found the vector for lambda=1. It could be [1 -1] ?
Exact same thing, it doesn't matter.
@@SeventhSolar thanks