- #1

karush

Gold Member

MHB

- 3,269

- 5

Find the general solution to the system of differential equations

$\begin{cases}

y'_1&=2y_1+y_2-y_3 \\

y'_2&=3y_2+y_3\\

y'_3&=3y_3

\end{cases}$

let

$y(t)=\begin{bmatrix}{y_1(t)\\y_2(t)\\y_3(t)}\end{bmatrix}

,\quad A=\begin{bmatrix}

2 & 1 & -1 \\

0 & 3 & 1 \\

0 & 0 & 3

\end{bmatrix}$

so

\begin{align*}\displaystyle

y_1&=c_1e^{2t}+c_2e^{t}+c_3e^{-t}\\

y_2&=c_2e^{3t}+c_3e^{t}\\

y_3&=c_3e^{3t}

\end{align*}

ok if correct so far assume next step is to diagonalize $A:\quad A=PDP^{-1}$

well according to EMH this is not diagonalizable but is look like a triangle

so would this be

$\begin{pmatrix} y'_1 \\ y'_2 \\ y'_3 \end{pmatrix}

= \begin{pmatrix} 2y_1&+y_2&-y_3 \\

0&3y_2&y_3\\

0&0&3y_3 \end{pmatrix}

\cdot

\begin{pmatrix} x \\ y \\ z \end{pmatrix}

+ \begin{pmatrix} b_1(t) \\ b_2(t) \\ b_3(t) \end{pmatrix}$

$\begin{cases}

y'_1&=2y_1+y_2-y_3 \\

y'_2&=3y_2+y_3\\

y'_3&=3y_3

\end{cases}$

let

$y(t)=\begin{bmatrix}{y_1(t)\\y_2(t)\\y_3(t)}\end{bmatrix}

,\quad A=\begin{bmatrix}

2 & 1 & -1 \\

0 & 3 & 1 \\

0 & 0 & 3

\end{bmatrix}$

so

\begin{align*}\displaystyle

y_1&=c_1e^{2t}+c_2e^{t}+c_3e^{-t}\\

y_2&=c_2e^{3t}+c_3e^{t}\\

y_3&=c_3e^{3t}

\end{align*}

ok if correct so far assume next step is to diagonalize $A:\quad A=PDP^{-1}$

well according to EMH this is not diagonalizable but is look like a triangle

so would this be

$\begin{pmatrix} y'_1 \\ y'_2 \\ y'_3 \end{pmatrix}

= \begin{pmatrix} 2y_1&+y_2&-y_3 \\

0&3y_2&y_3\\

0&0&3y_3 \end{pmatrix}

\cdot

\begin{pmatrix} x \\ y \\ z \end{pmatrix}

+ \begin{pmatrix} b_1(t) \\ b_2(t) \\ b_3(t) \end{pmatrix}$

Last edited: