Question

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 

\(\lambda1=3\Rightarrow \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\)

\(\lambda2=0\Rightarrow \begin{bmatrix} 1 \\ 5 \\ 1 \end{bmatrix}\begin{bmatrix}2 \\ 1 \\ 4 \end{bmatrix}\)

Answer 1

By theorem 6.19 we know that the solution is
\(y=c_1e^{\lambda_1t}u_1+\dotsc+c_ne^{\lambda_nt}u_n\)
with \(\lambda_i\) the eigenvalues of the matrix A nad u_i the eigenvectors Thus for this case we then obtain the general solution:
\(\begin{bmatrix}y_1\\ y_2\\ y_3 \end{bmatrix}=y = c_1e^{3t}\begin{bmatrix}1\\1\\0 \end{bmatrix}+c_2e^{0t} \begin{bmatrix}1\\5\\1 \end{bmatrix}+ c_3e^{0t}\begin{bmatrix}2\\1\\4 \end{bmatrix}\)
Thus we obtain:
\(y_1=c_1e^{3t}+c_2e^{0t}+2c_3e^{0t}=c_1e^{3t}+c_2+2c_3\)
\(y_2=c_1e^{3t}+5c_2e^{0t}+c_3e^{0t}=c_1e^{3t}+5c_2+c_3\)
\(y_3=c_2e^{0t}+4c_3e^{0t}=c_2+4c_3\)