Deangelo Hardy

2022-04-07

Find the solution of the system ${x}^{\prime}=x-2y\text{}\text{and}\text{}{y}^{\prime}=4x+5y$ .

Ausspruchx807

Beginner2022-04-08Added 6 answers

Step 1

The solution for such systems is as follows:

Let

$=\left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]$

We have ${\lambda}_{1}=3+2i,{\lambda}_{2}=\overline{{\lambda}_{1}}=3-2i$ and

${v}_{1}=\left[\begin{array}{c}-1+i\\ 2\end{array}\right]=\left[\begin{array}{c}-1\\ 2\end{array}\right]+i\left[\begin{array}{c}1\\ 0\end{array}\right]$

and ${v}_{2}=\left[\begin{array}{c}-1-i\\ 2\end{array}\right]=\left[\begin{array}{c}-1\\ 2\end{array}\right]+i\left[\begin{array}{c}-1\\ 0\end{array}\right]$

Step 2

Now we construct the matrix $P$ from the imaginary and real parts of the eigenvectors;

$P=\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]\Rightarrow {P}^{-1}=\frac{1}{13}\left[\begin{array}{cc}3& 2\\ -2& 3\end{array}\right]$

and

${P}^{-1}AP=\left[\begin{array}{cc}3& -2\\ 2& 3\end{array}\right]$

And the final solution

$\left[\begin{array}{c}x\left(t\right)\\ y\left(t\right)\end{array}\right]=P{e}^{3t}\left[\begin{array}{cc}\mathrm{cos}2t& -\mathrm{sin}2t\\ \mathrm{sin}2t& \mathrm{cos}2t\end{array}\right]{P}^{-1}\left[\begin{array}{c}{x}_{0}\\ {y}_{0}\end{array}\right]$

and

$\left[\begin{array}{c}x\left(t\right)\\ y\left(t\right)\end{array}\right]=\frac{{e}^{3t}}{13}\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]\left[\begin{array}{cc}\mathrm{cos}2t& -\mathrm{sin}2t\\ \mathrm{sin}2t& \mathrm{cos}2t\end{array}\right]\left[\begin{array}{cc}3& 2\\ -2& 3\end{array}\right]\left[\begin{array}{c}{x}_{0}\\ {y}_{0}\end{array}\right]$

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