Solving 4 Differential Equations - Tips for Exam

[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 please.

 

[Hurkyl]

Can you solve a pair of such equations?

 

[VietDao29]

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 please.

Please do not double post!
https://www.physicsforums.com/showthread.php?t=109061 in Precalculus Mathematics is enough!
But do you study this in precalculus by the way?

 

[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.

 

[Bimbar]

Is problem easy to you ?

 

[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?

 

[SGT]

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

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
$$e^{At} = I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!}+ ...$$