Abram Boyd

2022-06-19

My Problem is this given System of differential equations.
${y}_{1}^{\mathrm{\prime }}=5{y}_{1}+2{y}_{2}\phantom{\rule{0ex}{0ex}}{y}_{2}^{\mathrm{\prime }}=-2{y}_{1}+{y}_{2}$
My Approach was: again, i analyze, it must be a ordinary, linear System of equations, with both being of first-order. Than i built the corresponding Matrix as follows:
$\underset{{\stackrel{\to }{y}}^{\mathrm{\prime }}}{\underset{⏟}{\left(\begin{array}{c}{y}_{1}^{\mathrm{\prime }}\\ {y}_{2}^{\mathrm{\prime }}\end{array}\right)}}=\underset{\mathbf{A}}{\underset{⏟}{\left(\begin{array}{cc}5& 2\\ -2& 1\end{array}\right)}}\underset{\stackrel{\to }{y}}{\underset{⏟}{\left(\begin{array}{c}{y}_{1}\\ {y}_{2}\end{array}\right)}}$
that's why:
${\stackrel{\to }{y}}^{\mathrm{\prime }}=\left(\begin{array}{cc}5& 2\\ -2& 1\end{array}\right)\stackrel{\to }{y}$
Then I determined the eigenvalues:
they are ${r}_{1}=3$ and ${r}_{2}=3$
Knowing them, I can build the corresponding eigenvectors:
they are ${\stackrel{\to }{v}}_{1}=\left(\begin{array}{c}-1\\ +1\end{array}\right)$ and ${\stackrel{\to }{v}}_{2}=\left(\begin{array}{c}0\\ 0\end{array}\right)$
Now i plug into the equation:
$\stackrel{\to }{x}={c}_{1}{e}^{{r}_{1}t}\stackrel{\to }{{v}_{1}}+{c}_{2}{e}^{{r}_{2}t}\stackrel{\to }{{v}_{2}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{x}={c}_{1}{e}^{3t}\left(\begin{array}{c}-1\\ 1\end{array}\right)+{c}_{2}{e}^{3t}\left(\begin{array}{c}0\\ 0\end{array}\right)$
${y}_{1}=-{c}_{1}{e}^{3t}+0{c}_{2}{e}^{3t}\phantom{\rule{0ex}{0ex}}{y}_{2}={c}_{1}{e}^{3t}+0{c}_{2}{e}^{3t}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{y}_{1}=-{c}_{1}{e}^{3t}\phantom{\rule{0ex}{0ex}}{y}_{2}={c}_{1}{e}^{3t}$
But I doubt it's correct. My suspect are the eigenvectors, I really don't know if they are correct. And this could have lead to a wrong solution.

Mateo Barajas

Hint:
${X}^{\prime }\left(t\right)=AX\left(t\right)$
if you have repeated eigen value like $c$ and $v$ is eigen vector correspond to $c$ then general solution is
$X\left(t\right)={e}^{ct}v$
and here solution is:
$X\left(t\right)=c{e}^{3t}\left(\begin{array}{c}-1\\ 1\end{array}\right)$

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