MHB -16.1 Find the general solution to the system of DE

[karush]

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

$\begin{pmatrix} y'_1 \\ y'_2 \\ y'_3 \end{pmatrix}
= \begin{pmatrix} 2y_1&+y_2&-y_3 \\
0&3y_2&y_3\\
0&0&3y_3 \end{pmatrix}
\cdot
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
+ \begin{pmatrix} b_1(t) \\ b_2(t) \\ b_3(t) \end{pmatrix}$

 

[HOI]

What "next step"? The problem asked you to find y₁, y₂, and y₃ and you have already done that!

(Unfortunately you also have the wrong solution!)

 

[HallsofIvy]

Personally, I would not use "matrices" at all.

Find the general solution to the system of differential equations
$\begin{cases}
y'_1&=2y_1+y_2-y_3 \\
y'_2&=3y_2+y_3\\
y'_3&=3y_3
\end{cases}$
\

The last equation, $y'_3= 3y_3$ is a first order, linear, homogeneous equation with constant coefficients. It's "characteristic equation" is $r= 3$ so the general solution is $y= Ae^{3x}$. (You could also have written this as $\frac{dy}{y}= 3dx$ and integrate to get the same solution.)

Once you know that the second equation can be written as $y'_2= 3y_2+ Ae^{3x}$, another first order linear equation with constant coefficients but now it is not homogeneous. Its characteristic equation is again $r= 3$ and the general solution to the "associated homogeneous equation" is again $y= Be^{3x}$. With right side "$Ae^{3t}$" we would normally try a solution to the entire equation of the form "$e^{3x}$ but because that satisfies the homogenous equation we try, instead, $y= Pxe^{3x}$. Then $y'= Pe^{3x}+ 3Pxe^{3x}$ and $e^{3x}+ 3Pe^{3x}=

3Pxe^{3x}+ Ae^{3x}$ so $P= A$ and $y_2= Be^{3x}+ Axe^{3x}$.

Now we can write the first equation as
$y'_1= 2y_1+ A^{3x}- Be^{3x}- Axe^{3x}= 2y_1+ (A- B)e^{3x}- Axe^{3x}$ and can be solved in much the same way.
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