# 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
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\}

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

### Relevant Questions

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
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The coefficient matrix for a system of linear differential equations of the form $$y_1=Ay$$
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The coefficient matrix for a system of linear differential equations of the form $$\displaystyle{y}^{{{1}}}={A}_{{{y}}}$$ has the given eigenvalues and eigenspace bases. Find the general solution for the system.
$$\displaystyle{\left[\lambda_{{{1}}}=-{1}\Rightarrow\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}{0}{3}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},\lambda_{{{2}}}={3}{i}\Rightarrow\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}-{i}{1}+{i}{7}{i}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},\lambda_{{3}}=-{3}{i}\Rightarrow\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{2}+{i}{1}-{i}-{7}{i}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace}\right]}$$
The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution.
$$\displaystyle{\left[\begin{matrix}{1}&{0}&-{1}&-{2}\\{0}&{1}&{2}&{3}\end{matrix}\right]}$$
The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution.
$$\displaystyle{\left[\begin{matrix}{1}&-{1}&{0}&-{2}&{0}&{0}\\{0}&{0}&{1}&{2}&{0}&{0}\\{0}&{0}&{0}&{0}&{1}&{0}\end{matrix}\right]}$$
The reduced row echelon form of the augmented matrix of a system of linear equations is given. Determine whether this system of linear equations is consistent and, if so, find its general solution. Write the solution in vector form. $$\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}&−{1}&{3}&{9}\backslash{0}&{1}&{2}&−{5}&{8}\backslash{0}&{0}&{0}&{0}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}$$
The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the general solution. $$\begin{bmatrix}1&0&-1&-2&0\\0&1&2&3&0\end{bmatrix}$$
Write the homogeneous system of linear equations in the form AX = 0. Then verify by matrix multiplication that the given matrix X is a solution of the system for any real number $$c_1$$
$$\begin{cases}x_1+x_2+x_3+x_4=0\\-x_1+x_2-x_3+x_4=0\\ x_1+x_2-x_3-x_4=0\\3x_1+x_2+x_3-x_4=0 \end{cases}$$
$$X =\begin{pmatrix}1\\-1\\-1\\1\end{pmatrix}$$
Form (a) the coefficient matrix and (b) the augmented matrix for the system of linear equations $$\begin{cases}9x-3y+z=13 \\ 12x-8z=5 \\ 3x+4y-z =6 \end{cases}$$
$$\displaystyle{\left\lbrace\begin{matrix}{x}+{y}={0}\\{5}{x}-{2}{y}-{2}{z}={12}\\{2}{x}+{4}{y}+{z}={5}\end{matrix}\right.}$$