# Solving a system of differential equations using diagonalization

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.

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

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 $$y_1 = c_1e^t, y_12 = c_2e^{-4t} - \frac 25c_1e^t.$$
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.