I believe the standard method is to write the system as a matrix equation: d/dx [y] = [1 -5] [y] [z] [1 7] [z] Solve the characteristic equation chi(lambda) = | 1-lambda -5 | = lambda^2 - 8 lambda + 12 = 0 | 1 7-lambda | lambda = 2 or 6, eigenvalues. Skipping a few details... The corresponding eigenvectors are [y] = [ 5] or [ 1] [z] [ -1] [-1] Solutions are [y] = c1 e^(2x)[5 ] + c2 e^(6x) [ 1] [z] [-1] [-1] Of course the answer is equivalent, but the method is easier to extend to larger systems.
Another method you can use that will get you the same answer is with a matrix exponential. The system written in terms of matrices and vectors can be represented as X'=Ax, which resembles the traditional ode y'=ay. For the traditional differential equation since we know y=c*e^(ay) its almost natural to consider the solution to the system of odes to be X=c*e^(Ax). Using the Taylor expansion of e^x and some properties of matrix diagnolization its pretty easy to come to the same answer you have.
Yes, as an amateur I ask, how does someone know to use e^x to substitute for the variables in the system? Why couldn't sin(x) or something be used? Thanks for the video!
A reasonable question. Once you have got it to y'' - 8y' + 12y = 0 you can see that sin(x) or cos(x) won't help, as their derivatives give you the opposite one (-sin x in the case of cos x). Whereas, (e^x)' = e^x, which is the reason this function is so important.
Nice - that was fun to watch!
I believe the standard method is to write the system as a matrix equation:
d/dx [y] = [1 -5] [y]
[z] [1 7] [z]
Solve the characteristic equation
chi(lambda) = | 1-lambda -5 | = lambda^2 - 8 lambda + 12 = 0
| 1 7-lambda |
lambda = 2 or 6, eigenvalues.
Skipping a few details...
The corresponding eigenvectors are [y] = [ 5] or [ 1]
[z] [ -1] [-1]
Solutions are [y] = c1 e^(2x)[5 ] + c2 e^(6x) [ 1]
[z] [-1] [-1]
Of course the answer is equivalent, but the method is easier to extend to larger systems.
Another method you can use that will get you the same answer is with a matrix exponential. The system written in terms of matrices and vectors can be represented as X'=Ax, which resembles the traditional ode y'=ay. For the traditional differential equation since we know y=c*e^(ay) its almost natural to consider the solution to the system of odes to be X=c*e^(Ax). Using the Taylor expansion of e^x and some properties of matrix diagnolization its pretty easy to come to the same answer you have.
Shouldn't you be using ∂y/∂x etc ?
Yes, as an amateur I ask, how does someone know to use e^x to substitute for the variables in the system? Why couldn't sin(x) or something be used?
Thanks for the video!
A reasonable question. Once you have got it to
y'' - 8y' + 12y = 0
you can see that sin(x) or cos(x) won't help, as their derivatives give you the opposite one (-sin x in the case of cos x). Whereas, (e^x)' = e^x, which is the reason this function is so important.