# System of Differential equations using Laplace.

#### rellsun

##### New member
Hey everyone i have a system of Differential equations that must be solved using Laplace transformation thats giving me trouble.
Heres the problem :. all i have so far is this: x'=5x+5y+2z, y'=-6x-6y-5z, z'=6x+6y+5z.

#### CurveGuy

##### New member
x'=5x+5y+2z, y'=-6x-6y-5z, z'=6x+6y+5z.
Its been many years, but the Laplace( x') = s * Laplace( x) if my memory serves me correctly. Thus, rewrite your equations like this: sX = 5X + 5Y + 2 Z, sY = -6X - 6Y - 5Z, and so on. Now collect terms on the left, e.g.
(s - 5)X = 5Y + 2Z ==> X = (5Y + 2Z) / (s -5). Do this same thing for Y & Z. Now you have a system of Laplace equations. Solve this system of 's' dependent variables and you'll have an answer.

#### LCKurtz

##### Full Member
CurveGuy is correct about the method but assumed the initial conditions are all 0. For example $$\displaystyle \dot X = sX-x(0)$$. This leads to the linear system:

$$\displaystyle \begin{pmatrix} sX-x(0)\\sY-y(0)\\sZ-z(0) \end{pmatrix} = \begin{pmatrix} 5 & 5 & 2 \\ -6 & -6 & -5 \\ 6 & 6 & 5 \end{pmatrix} = \begin{pmatrix} X \\ Y \\ Z \end{pmatrix}$$
where the initial conditions might not be $$\displaystyle 0$$. You can solve the system by any favorite method you choose to get the transforms $$\displaystyle X,~Y,~Z$$ after which you need to find their inverses.
That $$\displaystyle 6$$ with a line above it should be $$\displaystyle -6$$. There appears to be a problem with Latex rendering.