Solving system of linear differential equations by eigenvalues
Using eigenvalues and eigenvectors solve system of differential equations:
$$x_1'=x_1+2x_2$$ $$x_2' = 2x_1+x_2$$
And find solution for the initial conditions: $x_1(0) = 1; x_2(0) = -1$
I tried to solve it, but I don't have right results, so I can't check my solution. I would like someone to write how he would solve it and what results would he get.
$\endgroup$ 34 Answers
$\begingroup$The matrix A of coefficients has two eigenvalues, $3$ and $-1$, and the corresponding eigenvectors $$v_1=(1,1)\quad v_2(1,-1)$$ If you write your system like: $Av=v'$ where $v=(x_1(t),x_2(t))$ you find that $v_1(t)=e^{3t}v_1$ and $v_2(t)=e^{-t}v_2$ you see that (the) solution is given by $v=c_1v_1(t)+c_2v_2(t)$. Then you impose the initial conditions. Sorry for the poor english.
$\endgroup$ $\begingroup$The system matrix is $$A=\begin{bmatrix}1 &2\\ 2& 1 \end{bmatrix}$$ The eigenvalues are $-1$ and $3$. $$\Lambda=\begin{bmatrix}-1 & 0\\0& 3\end{bmatrix}$$
The eigenvector matrix is (normalized columns) $$Q=\begin{bmatrix}-0.7071 & 0.7071\\ 0.7071 & 0.7071\end{bmatrix}=\begin{bmatrix}q_1 &q_2\end{bmatrix}$$ This matrix is orthogonal.
Then $x=[x_1 \quad x_2]'$:
$$x(t)=e^{At}x(0)=Qe^{\Lambda t}Q'x(0)=q_1q_1'x(0)e^{-t}+q_2q_2'x(0)e^{3t}$$
$\endgroup$ $\begingroup$i really need to get more reputation so i can comment. anyway heres my attempt. hope it helps.
you have $$\dot{x}=x+2y\\\dot{y}=2x+y$$
the systems linear so it can be rewritten as $$\vec{\dot{x}}=A\vec{x}$$ $$\vec{\dot{x}}=\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}$$
calculating $det[A-\lambda I]$ gives $det\begin{bmatrix} 1-\lambda & 2 \\ 2 & 1-\lambda \end{bmatrix}$ simplifing to $\lambda^{2}-2\lambda-3=0 \Rightarrow (\lambda-3)(\lambda+1)$
Calculate the eigan vectors now. $\begin{bmatrix} 1-3 & 2 \\ 2 & 1-3 \end{bmatrix}\begin{bmatrix} \vec{v_{1}}\\ \vec{v_{2}} \end{bmatrix}=0 \Rightarrow\begin{bmatrix} 1\\ 1 \end{bmatrix}$ and $\begin{bmatrix} 1--1 & 2 \\ 2 & 1--1 \end{bmatrix}\begin{bmatrix} \vec{v_{1}}\\ \vec{v_{2}} \end{bmatrix}=0 \Rightarrow\begin{bmatrix} 1\\ -1 \end{bmatrix}$ (sorry if someone could check this last one for me that'd be great)
This gives the solution $$\vec{x}=c_{1}e^{3t}\begin{bmatrix}1\\1\end{bmatrix}+c_{2}e^{-t}\begin{bmatrix}1\\-1\end{bmatrix}$$ substituting initial conditions give $$\vec{x(0)}=c_{1}\begin{bmatrix}1\\1\end{bmatrix}+c_{2}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}1\\-1\end{bmatrix}$$ implies $c_{1}=1 $ and $c_{2}=0$
giving final solution $$\vec{x}=e^{3t}\begin{bmatrix}1\\1\end{bmatrix}$$
$\endgroup$ 1 $\begingroup$Look here and here, and following their notation,$$A = \begin{pmatrix}
1 & 2 \\
2 & 1 \\
\end{pmatrix},$$ $$\lambda_1 = 3 \text{ and } \lambda_2 = -1,$$ $$\text{& } \vec{\eta}_1 = \begin{pmatrix}
1 \\
1 \\
\end{pmatrix} \text{ and } \vec{\eta}_2 = \begin{pmatrix}
1 \\
-1 \\
\end{pmatrix}.$$Then, $\vec{x}(t)=c_1e^{3t}\begin{pmatrix}
1 \\
1 \\
\end{pmatrix}+c_2e^{-t}\begin{pmatrix}
1 \\
-1 \\
\end{pmatrix} = \begin{pmatrix}
c_1e^{3t}+c_2e^{-t} \\
c_1e^{3t}-c_2e^{-t} \\
\end{pmatrix}.$
Therefore, $\begin{cases}
x_1=c_1e^{3t}+c_2e^{-t} \\
x_2=c_1e^{3t}-c_2e^{-t}.
\end{cases}$
Don't forget to solve for $c_1$ and $c_2$ as well.
$\endgroup$
