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:

$\underset{{\overrightarrow{y}}^{\mathrm{\prime}}}{\underset{\u23df}{\left(\begin{array}{c}{y}_{1}^{\mathrm{\prime}}\\ {y}_{2}^{\mathrm{\prime}}\end{array}\right)}}=\underset{\mathbf{A}}{\underset{\u23df}{\left(\begin{array}{cc}5& 2\\ -2& 1\end{array}\right)}}\underset{\overrightarrow{y}}{\underset{\u23df}{\left(\begin{array}{c}{y}_{1}\\ {y}_{2}\end{array}\right)}}$

that's why:

${\overrightarrow{y}}^{\mathrm{\prime}}=\left(\begin{array}{cc}5& 2\\ -2& 1\end{array}\right)\overrightarrow{y}$

Then I determined the eigenvalues:

they are ${r}_{1}=3$ and ${r}_{2}=3$

Knowing them, I can build the corresponding eigenvectors:

they are ${\overrightarrow{v}}_{1}=\left(\begin{array}{c}-1\\ +1\end{array}\right)$ and ${\overrightarrow{v}}_{2}=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Now i plug into the equation:

$\overrightarrow{x}={c}_{1}{e}^{{r}_{1}t}\overrightarrow{{v}_{1}}+{c}_{2}{e}^{{r}_{2}t}\overrightarrow{{v}_{2}}\phantom{\rule{0ex}{0ex}}\overrightarrow{x}={c}_{1}{e}^{3t}\left(\begin{array}{c}-1\\ 1\end{array}\right)+{c}_{2}{e}^{3t}\left(\begin{array}{c}0\\ 0\end{array}\right)$

this lead to my result:

${y}_{1}=-{c}_{1}{e}^{3t}+0{c}_{2}{e}^{3t}\phantom{\rule{0ex}{0ex}}{y}_{2}={c}_{1}{e}^{3t}+0{c}_{2}{e}^{3t}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{y}_{1}=-{c}_{1}{e}^{3t}\phantom{\rule{0ex}{0ex}}{y}_{2}={c}_{1}{e}^{3t}$

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.

${y}_{1}^{\mathrm{\prime}}=5{y}_{1}+2{y}_{2}\phantom{\rule{0ex}{0ex}}{y}_{2}^{\mathrm{\prime}}=-2{y}_{1}+{y}_{2}$

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:

$\underset{{\overrightarrow{y}}^{\mathrm{\prime}}}{\underset{\u23df}{\left(\begin{array}{c}{y}_{1}^{\mathrm{\prime}}\\ {y}_{2}^{\mathrm{\prime}}\end{array}\right)}}=\underset{\mathbf{A}}{\underset{\u23df}{\left(\begin{array}{cc}5& 2\\ -2& 1\end{array}\right)}}\underset{\overrightarrow{y}}{\underset{\u23df}{\left(\begin{array}{c}{y}_{1}\\ {y}_{2}\end{array}\right)}}$

that's why:

${\overrightarrow{y}}^{\mathrm{\prime}}=\left(\begin{array}{cc}5& 2\\ -2& 1\end{array}\right)\overrightarrow{y}$

Then I determined the eigenvalues:

they are ${r}_{1}=3$ and ${r}_{2}=3$

Knowing them, I can build the corresponding eigenvectors:

they are ${\overrightarrow{v}}_{1}=\left(\begin{array}{c}-1\\ +1\end{array}\right)$ and ${\overrightarrow{v}}_{2}=\left(\begin{array}{c}0\\ 0\end{array}\right)$

Now i plug into the equation:

$\overrightarrow{x}={c}_{1}{e}^{{r}_{1}t}\overrightarrow{{v}_{1}}+{c}_{2}{e}^{{r}_{2}t}\overrightarrow{{v}_{2}}\phantom{\rule{0ex}{0ex}}\overrightarrow{x}={c}_{1}{e}^{3t}\left(\begin{array}{c}-1\\ 1\end{array}\right)+{c}_{2}{e}^{3t}\left(\begin{array}{c}0\\ 0\end{array}\right)$

this lead to my result:

${y}_{1}=-{c}_{1}{e}^{3t}+0{c}_{2}{e}^{3t}\phantom{\rule{0ex}{0ex}}{y}_{2}={c}_{1}{e}^{3t}+0{c}_{2}{e}^{3t}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{y}_{1}=-{c}_{1}{e}^{3t}\phantom{\rule{0ex}{0ex}}{y}_{2}={c}_{1}{e}^{3t}$

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

Beginner2022-06-20Added 13 answers

Hint:

${X}^{\prime}(t)=AX(t)$

if you have repeated eigen value like $c$ and $v$ is eigen vector correspond to $c$ then general solution is

$X(t)={e}^{ct}v$

and here solution is:

$X(t)=c{e}^{3t}\left(\begin{array}{c}-1\\ 1\end{array}\right)$

${X}^{\prime}(t)=AX(t)$

if you have repeated eigen value like $c$ and $v$ is eigen vector correspond to $c$ then general solution is

$X(t)={e}^{ct}v$

and here solution is:

$X(t)=c{e}^{3t}\left(\begin{array}{c}-1\\ 1\end{array}\right)$