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16.1 Find the general solution to the system of differential equations

karush

Well-known member
Jan 31, 2012
2,886
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

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

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
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!)
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.