# The coefficient matrix for a system of linear differential equations of the form y^1=Ay

Forms of linear equations

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

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

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

2021-01-05
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:
$$y_1=c_1e^{3t}+c_2e^{0t}+2c_3e^{0t}=c_1e^{3t}+c_2+2c_3$$
$$y_2=c_1e^{3t}+5c_2e^{0t}+c_3e^{0t}=c_1e^{3t}+5c_2+c_3$$
$$y_3=c_2e^{0t}+4c_3e^{0t}=c_2+4c_3$$