Solution to system of differential equations

[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

 

[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\}

 

[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.