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29.1 give a system of fundamental solutions

karush

Well-known member
Jan 31, 2012
2,886
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
Feb 7, 2012
2,770
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
Jan 31, 2012
2,886
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
Jan 26, 2012
4,198
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
Jan 31, 2012
2,886
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
Jan 26, 2012
4,198
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.