5
$\begingroup$

My Problem is this given System of differential Equations: $$\dot{x}=8x+18y$$ $$\dot{y}=-3x-7y$$ I am looking for a gerenal solution.

My Approach was: i can see this is a System of linear and ordinary differential equations. Both are of first-order, because the highest derivative is the first. But now i am stuck, i have no idea how to solve it. A Transformation into a Matrix should lead to this expression: $$\overrightarrow{y}=\left( \begin{array}{cc} 8 & 18 \\ -3 & -7 \end{array} \right)\cdot x$$ or is this correct: $$\overrightarrow{x}=\left( \begin{array}{cc} 8 & 18 \\ -3 & -7 \end{array} \right)\cdot y\text{ ?}$$

But i don't know how to determine the solution, from this point on.

$\endgroup$

2 Answers 2

Reset to default
6
$\begingroup$

I'm going to rename your variables. Instead of $x$ and $y$, I will use $x_1$ and $x_2$ (respectively).

Now, let's look at the system:

$$\begin{cases} \dot x_1 = 8x_1+18x_2\\ \dot x_2 = -3x_1 -7x_2 \end{cases}$$

To change this into matrix form, we rewrite as $\dot {\vec x} = \mathbf A \vec x$, where $\mathbf A$ is a matrix.

This looks like: $$\underbrace{\pmatrix{\dot x_1 \\ \dot x_2}}_{\large{\dot {\vec x}}} = \underbrace{\pmatrix{8 & 18 \\ -3 & -7}}_{\large{\mathbf A}}\underbrace{\pmatrix{x_1\\x_2}}_{\large{\vec x}}$$

To solve the system, we find the eigenvalues of the matrix. These are $r_1 = 2$ and $r_2 = -1$. Two corresponding eigenvectors are $\vec \xi_1 =\pmatrix{3 \\ -1}$ and $\vec \xi_2 =\pmatrix{2 \\ -1}$, respectively.

We now plug these into the equation: $$\vec{x} = c_1e^{r_1t}\vec{\xi_1}+c_2e^{r_2t}\vec{\xi_2}$$

This yields: $$\vec{x} = c_1e^{2t}\pmatrix{3 \\ -1}+c_2e^{-t}\pmatrix{2 \\ -1}$$

So, your individual solutions are: $$x_1 = 3c_1e^{2t} + 2c_2e^{-t}\\ x_2 = -c_1e^{2t} -c_2e^{-t}$$

$\endgroup$
1
  • $\begingroup$ Nice write up +1 $\endgroup$ Commented Jun 30, 2013 at 20:29
3
$\begingroup$

I think you want this matrix:

$$\left( \begin{array}{cc} \dot{x} \\ \dot{y} \end{array}\right)=\left( \begin{array}{cc} 8 & 18 \\ -3 & -7 \end{array} \right)\cdot \left( \begin{array}{cc} x \\ y \end{array}\right),$$

and you then diagonalize the coupling matrix to get decoupled equations.

Note, you have something like this: $$\dot{X}=M\cdot X,$$ where: $$X=\left( \begin{array}{cc} x \\ y \end{array}\right),\,\dot{X}=\left( \begin{array}{cc} \dot{x} \\ \dot{y} \end{array}\right).$$

Then if the matrix $M$ is diagonalizable (this one is) you can write it as: $$M=S\cdot D \cdot S^{-1},$$ where $D$ is a diagonal matrix. You can then manipulate the differential equation as follows: $$\dot{X}=S\cdot D \cdot S^{-1}\cdot X,$$ $$S^{-1}\cdot\dot{X}=D \cdot S^{-1}\cdot X,$$ $$\dot{U}=D \cdot U,$$ where $U=S^{-1}\cdot X$. This then gives you two decoupled differential equations to solve:

$$\dot{U_1}=D_1U_1.$$ $$\dot{U_2}=D_2U_2.$$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.