Burhan Hopper

2021-01-04

The coefficient matrix for a system of linear differential equations of the form ${y}^{1}=Ay$  has the given eigenvalues and eigenspace bases. Find the general solution for the system

$\lambda 1=3⇒\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$

$\lambda 2=0⇒\left[\begin{array}{c}1\\ 5\\ 1\end{array}\right]\left[\begin{array}{c}2\\ 1\\ 4\end{array}\right]$

Bella

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:
${y}_{1}={c}_{1}{e}^{3t}+{c}_{2}{e}^{0t}+2{c}_{3}{e}^{0t}={c}_{1}{e}^{3t}+{c}_{2}+2{c}_{3}$
${y}_{2}={c}_{1}{e}^{3t}+5{c}_{2}{e}^{0t}+{c}_{3}{e}^{0t}={c}_{1}{e}^{3t}+5{c}_{2}+{c}_{3}$
${y}_{3}={c}_{2}{e}^{0t}+4{c}_{3}{e}^{0t}={c}_{2}+4{c}_{3}$

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