Burhan Hopper

2021-01-04

The coefficient matrix for a system of linear differential equations of the form

Bella

Skilled2021-01-05Added 81 answers

By theorem 6.19 we know that the solution is

$y={c}_{1}{e}^{{\lambda}_{1}t}{u}_{1}+\dots +{c}_{n}{e}^{{\lambda}_{n}t}{u}_{n}$

with${\lambda}_{i}$ the eigenvalues of the matrix A nad u_i the eigenvectors
Thus for this case we then obtain the general solution:

$\left[\begin{array}{c}{y}_{1}\\ {y}_{2}\\ {y}_{3}\end{array}\right]=y={c}_{1}{e}^{3t}\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]+{c}_{2}{e}^{0t}\left[\begin{array}{c}1\\ 5\\ 1\end{array}\right]+{c}_{3}{e}^{0t}\left[\begin{array}{c}2\\ 1\\ 4\end{array}\right]$

Thus we obtain:

${y}_{1}={c}_{1}{e}^{3t}+{c}_{2}{e}^{0t}+2{c}_{3}{e}^{0t}={c}_{1}{e}^{3t}+{c}_{2}+2{c}_{3}$

${y}_{2}={c}_{1}{e}^{3t}+5{c}_{2}{e}^{0t}+{c}_{3}{e}^{0t}={c}_{1}{e}^{3t}+5{c}_{2}+{c}_{3}$

${y}_{3}={c}_{2}{e}^{0t}+4{c}_{3}{e}^{0t}={c}_{2}+4{c}_{3}$

with

Thus we obtain:

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