# 29.1 give a system of fundamental solutions

#### karush

##### Well-known member
determine their general solution and give a system of fundamental solutions.
use the different techniques of diagonal, diagonalizedable, or triangular.
$\begin{cases} y'_1 & =3y_1 \\ y'_2 & =2y_2\end{cases}$
set matrix
$A= \begin{pmatrix}0 &3\\0 &2\end{pmatrix}$
then find eigenvalues

ok just seeing if i am starting out correctly with this..
not sure what the difference is between diagonal and diagonalizedable? but it looks diagonal

#### Opalg

##### MHB Oldtimer
Staff member
determine their general solution and give a system of fundamental solutions.
use the different techniques of diagonal, diagonalizedable, or triangular.
$\begin{cases} y'_1 & =3y_1 \\ y'_2 & =2y_2\end{cases}$
set matrix
$A= \begin{pmatrix}0 &3\\0 &2\end{pmatrix}$
then find eigenvalues

ok just seeing if i am starting out correctly with this..
not sure what the difference is between diagonal and diagonalizedable? but it looks diagonal
It doesn't look diagonal to me, because of the $3$ that is off the diagonal. But it does look triangular.

#### karush

##### Well-known member
It doesn't look diagonal to me, because of the $3$ that is off the diagonal. But it does look triangular.
ok I don't know for sure but doesn't a diagonal have the property
 $A=PDP^{-1}$

which I think it does

#### Ackbach

##### Indicium Physicus
Staff member
Your DE doesn't match the matrix $A.$ Your system of DE's is actually de-coupled, as you've written it; for such a system, I would expect $A$ to be diagonal. So the matrix $A$ which matches your system would be
$$A=\left[\begin{matrix}3 &0 \\ 0 &2\end{matrix}\right].$$

#### karush

##### Well-known member
how would that fit to $y_1$ and $y_2$ since there is no $x_i$

not sure what you mean by decoupled

#### Ackbach

##### Indicium Physicus
Staff member
A system of equations like yours:
\begin{align*}
y_1'&=3y_1 \\
y_2'&=2y_2
\end{align*}
is very often written (with an eye towards using matrix methods) in the form $\mathbf{y}'=A\mathbf{y},$ where $A$ is a $2\times 2$ matrix, and
$$\mathbf{y}=\left[\begin{matrix}y_1\\y_2\end{matrix}\right].$$
So, if you compare $\mathbf{y}'=A\mathbf{y}$ with your system, it appears that $A$ must be
$$A=\left[\begin{matrix}3&0\\0&2\end{matrix}\right].$$
The lack of $x_i$ is not a problem here.

The system is called "decoupled" when the dependent variables don't show up in each others' DE's. So, in the equation $y_1'=3y_1,$ there's no $y_2,$ and in the equation $y_2'=2y_2,$ there's no $y_1.$ De-coupled DE's are very often considerably simpler to solve, since you can basically solve each one separately. Indeed, you can simply write down by inspection the solution to this system:
\begin{align*}
y_1&=C_1 e^{3x}\\
y_2&=C_2 e^{2x},
\end{align*}
assuming $x$ is the independent variable.