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

Staff Emeritus
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

Staff Emeritus
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