Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 6 down vote accepted

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

share|improve this answer
    
Nice write up +1 –  Amzoti Jun 30 '13 at 20:29

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:

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.