The coefficient matrix for a system of linear differential equations of the form y^1=Ay =Ay has the given eigenvalues and eigenspace bases. Find the general solution for the system lambda_1=3 Rightarrow left{ begin{bmatrix}110 end{bmatrix} right} , lambda_2=0 Rightarrow left{ begin{bmatrix}151 end{bmatrix} , begin{bmatrix}214 end{bmatrix} right}

Question
Forms of linear equations
asked 2021-01-04
The coefficient matrix for a system of linear differential equations of the form \(y^1=Ay\)
=Ay has the given eigenvalues and eigenspace bases. Find the general solution for the system \lambda_1=3 \Rightarrow \left\{ \begin{bmatrix}1\\1\\0 \end{bmatrix} \right\} , \lambda_2=0 \Rightarrow \left\{ \begin{bmatrix}1\\5\\1 \end{bmatrix} , \begin{bmatrix}2\\1\\4 \end{bmatrix} \right\}

Answers (1)

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:
\(y_1=c_1e^{3t}+c_2e^{0t}+2c_3e^{0t}=c_1e^{3t}+c_2+2c_3\)
\(y_2=c_1e^{3t}+5c_2e^{0t}+c_3e^{0t}=c_1e^{3t}+5c_2+c_3\)
\(y_3=c_2e^{0t}+4c_3e^{0t}=c_2+4c_3\)
0

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