Question

dY/dt = Ay, The derivative is given by this transformation matrix of A on the vector Y.
1.look for particular solutions by separting the variable
y(t)=e^(vt)x : Ax=Vx
2.Find all eigenvalues and eigenvectors of A:
Au_j=V_j*U_j, j=1,2,3...n

3.if eigen vectors {u1,u2,u3..un} are linearly independent(form a basis in R^n)
y(t)=c1e^(V_1*t)+c2e^(V_2*t)+...+cne^(V_n*t)

Answer 1

okay i think i do follow it now, but yes do write it out, ill be nice to see the steps laid out

Answer 2

So you want to understand why the matrix thingy works? Like at what point does it stop making sense or seem unreasonable to you?

Like in the past when we had \[\LARGE y'' + 3y' +2y=0\] we could just use \[\LARGE y=e^{r t}\] and solve \[\LARGE r^2+3r+2r=(r+1)(r+2)\] to get \[\LARGE y=C_1e^{-t}+C_2e^{-2t}\] It's basically just an extension of this.

Answer 3

I am not really sure I know how to formally prove it, but the matrix is just a quick shorthand for doing a bunch of tedious calculations it's not like we're really doing something crazy.

Answer 4

yeah

Answer 5

Just looking at a single term, you get \[\LARGE y_n=e^{\lambda_n t}\] so the derivative is \[\LARGE y_n'=\lambda_n e^{\lambda_n t}\] so we can really consider this as a scalar multiple. \[\LARGE \frac{d}{dt}y_n=\lambda_n y_n\] where lambda is the eigen value and d/dt is the transformation. That's why we call these eigen functions

Answer 6

okay this is what i got so far