# Solution to system of differential equations

1. Apr 22, 2006

### Alteran

In order to find streamlines for vector field, we need to solve this system of differential equations:

$$\frac{dx}{dt} = -x+y+z$$
$$\frac{dy}{dt} = x-y+z$$
$$\frac{dz}{dt} = x+y-z$$
where
$$x(0) = a$$
$$y(0) = b$$
$$z(0) = c$$

I have used Mathematica in order to find solutions for these equations and it calculated the following:
$$x = \frac{2a-b-c+ae^{3t}+be^{3t}+ce^{3t}}{3e^{2t}}$$
$$y = \frac{-a+2b-c+ae^{3t}+be^{3t}+ce^{3t}}{3e^{2t}}$$
$$z = \frac{-a-b+2c+ae^{3t}+be^{3t}+ce^{3t}}{3e^{2t}}$$

it is, of course, right - I have check parametric plot and it is streamline, but I think how it came to that solution? Does anybody can explain how I can solve these equations (step-by-step would be very very helpful :roll: ) without using of Mathematica? I do not understand why there is $$3e^{2t}$$?

Thank you

2. Apr 22, 2006

### araven

The simpliest solution can be found using Laplace transform. The system you mentioned can be rewriten in a matrix form as following:

$$\frac{{d\bar x}}{{dt}} = A\bar x$$

where

$$\bar x = \left( {x,y,z} \right)$$

Applying Laplace transform to the both sides of the equation yields:

$$s\bar x\left( s \right) - \bar x_0 = A\bar x\left( s \right)$$

After rearranging and taking out the common terms you get:

$$\bar x\left( s \right) = \left( {sI - A} \right)^{ - 1} \bar x_0$$

Last, you need to perform the inverse Laplace transform (this is where the $$3e^{2t}$$ is coming from):

$$\bar x\left( t \right) = L^{ - 1} \left\{ {\bar x\left( s \right)} \right\}$$

3. Apr 22, 2006

### Alteran

As I understand we derive the matrix from the system, then we calculate det(A-xI) and we get cubic equation and then roots: 1, -2, -2 which are eigenvalues.

I got an equation: $$-x^3-3x^2+4 = 0$$ so roots are indeed: 1, -2, -2.

How to create eigenvectors? I belief that we need so called eigenvector to get linearly independent solutions.