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Solve the system \begin{align*} y'_1&= \phantom{-2}y_1 \\ y'_2&= -2y_1-4y_2 \end{align*}

I think they want me to solve it by using diagonalization. So far so good. I got the following:

The coefficient matrix for the system is

$$A= \begin{bmatrix} \phantom{-}1 & \phantom{-}0 \\ -2 & -4 \end{bmatrix} $$

Then I find the eigenvalues

$$ \text{det}(A-\lambda I)= \begin{bmatrix} \phantom{-}1-\lambda & \phantom{-}0 \\ -2 & -4-\lambda \end{bmatrix}= (1-\lambda)(-4-\lambda) $$

The eigenvalues are $\lambda=1$ and $\lambda=-4$.

Then I will find the matrix $P$ that $A$ is diagonalized by. $P$ is made up by the eigenvectors of $A$. So I find the eigenvectors.

For $\lambda=1$

$$ \begin{bmatrix} \phantom{-}0 & \phantom{-}0 \\ -2 & -5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

and for $\lambda=-4$

$$ \begin{bmatrix} \phantom{-}5 & 0 \\ -2 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

Then I got stuck. What are the eigenvectors? I also need hints for the rest.

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  • $\begingroup$ Do you want to solve the system of equation only by matrix method ? OR other methods are acceptable ? $\endgroup$
    – Empty
    Commented Apr 3, 2016 at 19:16
  • $\begingroup$ Possible duplicate of Getting equation from differential equations $\endgroup$
    – flawr
    Commented Apr 3, 2016 at 19:19
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    $\begingroup$ Just solve the equations... The eigenvectors are $(5,-2)$ and $(0,1)$. What's the problem, really? $\endgroup$ Commented Apr 3, 2016 at 19:58
  • $\begingroup$ "I think they want me to solve it by using diagonalization": why ? $\endgroup$
    – user65203
    Commented May 8, 2018 at 7:15

2 Answers 2

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i think you may not want to solve this equation by diagonalising it. this system is decoupled, therefore we can solve it directly. you have $y_1 = c_1e^t.$ subbing this in the second equation gives you, $$y_2' + 4y_2 = -2c_1e^t \tag 1$$ now $(1)$ has $y_2 = c_2e^{-4t}$ for a homogenous solution and $-\frac 25c_1e^t$ for a particular solution. therefore the solution to the system of equation is $$y_1 = c_1e^t, y_12 = c_2e^{-4t} - \frac 25c_1e^t.$$

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  • $\begingroup$ So this means that the coefficient matrix cannot be diagonalized? $\endgroup$ Commented Apr 3, 2016 at 19:24
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    $\begingroup$ @AlimTeacher, no it does not; what it means is it is not necessary. this is like solving lower/upper triangular system of linear equations. $\endgroup$
    – abel
    Commented Apr 3, 2016 at 19:25
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To find an eigenvector corresponding to $\lambda = 1$, you need to determine values for $x_1$ and $x_2$ so that $0x_1+0x_2=0$ and $-2x_1-5x_2=0$. One solution to this is $x_1=5, x_2=-2$, so you can use $(5, -2)$ as your eigenvector.

You can perform a similar calculation to get an eigenvector corresponding to $\lambda=4$. Looking at $5x_1+0x_2=0$ and $-2x_1+0x_2=0$, we see that $(0,1)$ is an eigenvector.

This means that $\vec{y_1}=(5,-2)e^{t}$ is a solution to your system, and so is $\vec{y_2}=(0,1)e^{-4t}$. Since the two eigenvectors here are linearly independent, solutions to this system are linear combinations of $\vec{y_1}$ and $\vec{y_2}$. Thus, the general solution to the system is: $$\vec{y}=C_1(5,-2)e^t+C_2(0,1)e^{-4t},$$ where $C_1$ and $C_2$ are arbitrary constants.

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