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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.

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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: $$x_1 = 3c_1e^{2t} + 2c_2e^{-t}\\ x_2 = -c_1e^{2t} -c_2e^{-t}$$

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  • $\begingroup$ Nice write up +1 $\endgroup$ – Amzoti Jun 30 '13 at 20:29
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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:

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

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