# Equation solving

1. Feb 2, 2006

### Bimbar

I have 4 difftiate equations now I want to solve them but i have no idea
dx/dt=ax+by
dy/dt=cy+dz
dz/dt=ez+fu
du/dt=gu+hx

Given that a,b,c,d,e,f,g,h are constants.
x,y,z,u are functions(t)
This problme will appraer in thenext exam, I am sure, my taecher emphasized it many times .
Hitn me plz.

2. Feb 2, 2006

### Hurkyl

Staff Emeritus
Can you solve a pair of such equations?

3. Feb 2, 2006

### VietDao29

Please do not double post!!! :grumpy:
One post in Precalculus Mathematics is enough!
But do you study this in precalculus by the way?

4. Feb 2, 2006

### HallsofIvy

4 gets a bit complicated!

A way to get a handle on them is to use "operator" notation. Replace the derivative by the symbol "D" (for derivative of course!) :
Dx=ax+by
Dy=cy+dz
Dz=ez+fu
Du=gu+hx
and treat the "D" as if it were a constant (as long as you are dealing with "linear equations with constant coefficients" that works!) and solve the equations for x, y, z, u: the result will involve powers of D. Replace the D by the derivative again (i.e. Dx= dx/dt, D2x= d2/dt, etc.) and solve the resulting differential equations in a single function.

5. Feb 2, 2006

### Bimbar

Is problem easy to you ?

6. Feb 3, 2006

### HallsofIvy

IF I were given specific numbers in the four equations, yes, it would be easy for me. If I were required to write a general solution including the coefficients, a- h, it would be tedious be nothing especially difficult.

Hurkyl asked before, "Can you solve a pair of such equations?". In other words is just that there are so many equations or do you not understand the concepts involved?

7. Feb 3, 2006

### SGT

You can write it in matrix form:
$$\left[ \begin{array}{cc} \frac{dx}{dt}\\ \frac{dy}{dt}\\ \frac{dz}{dt}\\ \frac{du}{dt} \end{array} \right] = \left[ \begin{array}{cccc} a & b & 0 & 0\\ 0 & c & d & 0\\ 0 & 0 & e & f\\ h & 0 & 0 & g \end{array} \right] \cdot \left[ \begin{array}{cc} x\\ y\\ z\\ u \end{array} \right]$$

or

$$\vec {\frac{dv}{dt}} = A\cdot\vec{v}$$

The solution of the scalar equation:

$$\frac{dv}{dt} = av$$
is
$$v=e^{at}\cdot v_0$$

Similarly, the solution of the matrix differential equation is:
$$\vec{v} = e^{At}\cdot \vec{v_0}$$
where
$$e^{At} = I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!}+ ...$$

Last edited by a moderator: Feb 3, 2006