An Interesting System of Differential Equations

[@pwmiles56]

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.

[@scottleung9587]

Nice - that was fun to watch!

[@Nirmy001]

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.

[@peterburkhard8700]

Matrix method with wife vectors is simplest method. Eigenvalues are 2 and 6. Put solution in standard exponential form with undetermined coefficients for general solution.

[@johns.8246]

I prefer the eigenvector method. And why did you never bother to tackle y^3-y=x^3-x from previous videos? All you did was the y^3+y=x^3+x case which resulted in the boring y=x graph.

[@ytodolodemasVal]

Add the 2 equations and you Will get a simple equation for f=y+z

[@peterburkhard8700]

Eigenvectors

[@arthurcoward6979]

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!

[@pwmiles56]

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.

[@neuralwarp]

Shouldn't you be using ∂y/∂x etc ?

[@Risu0chan]

No, y(x) is a function of the single variable x. And z(x) also depends on x only. You use ∂ (partial d) when a function depends on two (or more) variables. Here we have two functions, but only one variable.