Solve $\frac{d\textbf{y}}{dt}=B\textbf{y}$
If B was Diagonal matrix it would be piece of cake but now I got this three equations
$\frac{dy_1}{dt}=y_1$
$\frac{dy_2}{dt}=2y_2-3y_3$
$\frac{dy_3}{dt}=3y_2+2y_3$
While I can solve the first one by just splitting the variables, I don't know how to solve the other two:
$y_1=Aexp(t)$
$y_2=?$, $y_3=?$
Thanks for any help.
On
For a hand calculation, using Laplace Transform (LT) is an alternative.
(thanks to @Doug M for the spotting of a computational error).
Let $a=y_1(0), b=y_2(0), c=y_3(0).$
Let us make a separate treatment of the LT of the first equation on one hand, and the second and third on the other hand.
$sY_1-a=Y_1 \ \iff \ Y_1=\dfrac{a}{s-1} \ \iff \ y_1(t)=ae^{t}$
$\begin{cases}sY_2-b&=&2Y_2-3Y_3\\sY_2-c&=&3Y_2+2Y_3\end{cases} \ \iff \ \begin{cases}(2-s)Y_2-3Y_3&=&-b\\3Y_2+(2-s)Y_3&=& \ \ c\end{cases} $
Solving this parametric system by Cramer (determinant) method, with $\Delta=(s-2)^2+3^2$, one obtains:
$$\begin{cases}Y_2&=&\dfrac{b(s-2)+3c}{(s-2)^2+3^2}\\Y_3&=&\dfrac{-c(s-2)+3b}{(s-2)^2+3^2} \end{cases},$$
from which it is immediate to obtain, using a table of LT, the result.
$$\begin{cases}y_2(t)&=&e^{2t}(b\cos(3t)+c\sin(3t))\\y_3(t)&=&e^{2t}(-c\cos(3t)-b\sin(3t)) \end{cases}$$
$\mathbf y' = B\mathbf y\\ \mathbf y = e^{Bt}\mathbf y(0)$
$e^{Bt} = \sum \frac {B^nt^n}{n!}$
If $B = P^{-1}DP$ then $B^n = P^{-1}D^nP$
$e^{Bt} = P^{-1} e^{Dt} P\\ e^{Dt} = \begin{bmatrix} e^{\lambda_1 t}\\&e^{\lambda_2 t}\\&&e^{\lambda_3 t}\end{bmatrix}$
Sometimes it is easier to work with:
$P\mathbf y = e^{Bt}(P\mathbf y(0))$
$y_1 = C_1 e^t\\ y_2 = C_2 e^{2t}\cos 3t + C_3 e^{2t}\sin 3t\\ y_3 = C_3 e^{2t}\cos 3t - C_2 e^{2t}\sin 3t$
Alternatively:
$y_2' = 2y_2 - 3y_3\\ y_2'' = 2y_2' - 3y_3'\\ y_2'' = 2(2y_2 - 3y_3) - 3(3y_2 + 2y_3)\\ y_2'' = -5 y_2 - 12 y_3\\ y_3 = \frac 13(2y_2-y_2')\\ y_2'' = -5y_2 - 8y_2 + 4y_2'\\ y_2'' - 4y_2' + 13y_2 = 0$
and that is a 2nd order diff eq you should know how to solve