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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.

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What is your solution? How did you get it? –  Ron Gordon Jan 22 '13 at 18:06
    
Let $$A := \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$$. Then $x' = Ax$ such that $x = \exp(tA)x_0$ where $x = (x_1,x_2)^T$,$x' = (x_1',x_2')^T$ and $x_0 = (1,-1)^T$. Compute the eigenvalues and eigenvectors to write $A = V \Lambda V^{-1}$ such that $\exp(tA) = V \exp(t \Lambda)V^{-1}$ –  André Jan 22 '13 at 18:11
    
my solution for initial conditions is $x=e^{-x}(1,-1)^T$ –  user50222 Jan 22 '13 at 18:13

3 Answers 3

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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.

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hint:solution of this system is $x(t)=(exp(at))v_1+(exp(bt))v_2$ that a is eigen value correspond to $v_1$ and b is eigen value correspond to $v_2$ for $A := \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ $v_1$,$v_2$ are eigen vector and second way :$x(t)=exp(At)x_0 $ you should calculate $exp(At)=\sum_{i=0}^{∞}$\fraq{A^i}{k!}$

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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 $x=[x_1 \quad x_2]'$:

$$x(t)=e^{At}x(0)=Qe^{\Lambda t}Q'x(0)=q_1q_1'x(0)e^{-t}+q_2q_2'x(0)e^{3t}$$

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