Thread: Differential Equation Systems

Hi, first time posting here, I am an engineering student - really stuck on part of differential systems, could someone help explain how to go about converting a system of n 1st order Differential Equations into a higher nth order differential equation,

Im working on a dynamics modelling problem involving fluid flow and have 4 1st order equations to convert into a single 4th order DE,

would appreciate any help, cheers

Hi there!

All you do is just algebra

Consider the following system of 2 equations (linear, inhomogeneous DEs)

$\displaystyle \dot x=ax+by+e$
$\displaystyle \dot y=cx+dy+f$

where $\displaystyle x\rightarrow x(t)$ and $\displaystyle y\rightarrow y(t)$

rewrite the system using differentials:

$\displaystyle \frac{dx}{dt}=ax+by+e$
$\displaystyle \frac{dy}{dt}=cx+dy+f$

Now, let's divide the second equation by the first, in order to eliminate the independent variable t:

$\displaystyle \frac{dy}{dt}\frac{dt}{dx}=\frac{cx+dy+f}{ax+by+e}$

or

$\displaystyle \frac{dy}{dx}=y'(x)=\frac{cx+dy+f}{ax+by+e}$

Now, we have a new, 1st order DE, where $\displaystyle y\rightarrow y(x)$ is the dependent and 'x' is the independent variable

$\displaystyle y'=\frac{cx+dy+f}{ax+by+e}$

***

But I suspect you meant something different:

Let's take our system again:

$\displaystyle \dot x=ax+by+e$
$\displaystyle \dot y=cx+dy+f$

Now, we take the first equation and:

(1). Differentiate it with respect to 't': $\displaystyle \ddot x=a\dot x+b\dot y$

(2). Express y with x: $\displaystyle y=\frac{\dot x-ax-e}{b}$

What we do now is gradually plug the second equation of the system into (1) and then plug (2) in the resulting equation:

$\displaystyle \ddot x=a\dot x+b\dot y=a\dot x+b(cx+dy+f)=a\dot x+bcx+dby+bf$

and

$\displaystyle \ddot x=a\dot x+bcx+db(\frac{\dot x-ax-e}{b})+bf$

$\displaystyle \ddot x=a\dot x+bcx+d(\dot x-ax-e)+bf$
$\displaystyle \ddot x=a\dot x+bcx+d\dot x-dax-ed+bf$

let's rearrange a bit:

$\displaystyle \ddot x-(a+d)\dot x+(ad-bc)x=bf-de$

this should be the general case of inhomogeneous 2nd order system. You can try it with the 4rth order system

now, as I look at it once again I noticed following:

Our system can be written as follows:

$\displaystyle \dot{\vec x}=A\vec x+\vec v$

The left hand side:

$\displaystyle \ddot x-(a+d)\dot x+(ad-bc)x$

is nothing more that the characteristic equation of the matrix of the coefficients in front of 'x' and 'y' of the system:

$\displaystyle (a+d)$ is $\displaystyle spur(A)$ and $\displaystyle (ad-bc)$ is $\displaystyle det(A)$

What is more, if you choose 'e' and 'f' to be functions of 't', they appear only in the right hand side of the 2nd order DE, so the inhomogeneity is conserved.

Hi Marine, thanks for that

I was wondering for a system of 4 DE's would you combine two of them at a time creating 2 2nd order equations, and then combine those two to create a single 4th order equation?

the equations look something like this

dH1/dt = aM - bH1
dH2/dt = cH1 - dH2 - dH3
dH3/dt = eH2 - fH3 +gH4
dH4/dt = jH3 - kH4

H 1:4 are the water levels in 4 connected tanks, M is the water flow into the first tank, and all the lower case letters represent constant coefficients, Im trying to find a 4th order equation for H4 = ?

Hi there!

Let's latex the system:

$\displaystyle \dot H_1= aM - bH_1$
$\displaystyle \dot H_2= cH_1 - dH_2 - dH_3$
$\displaystyle \dot H_3= eH_2 - fH_3 +gH_4$
$\displaystyle \dot H_4= jH_3 - kH_4$

the first equation I'll take out of the system (for now) and we define:$\displaystyle cH_1=P$

$\displaystyle \dot H_2=-dH_2- dH_3+ P$
$\displaystyle \dot H_3= eH_2 - fH_3 +gH_4$
$\displaystyle \dot H_4= jH_3 - kH_4$

This system is of order 3, so we differentiate the last equation 2 more times:

$\displaystyle H_4'''= jH_3'' - kH_4''$

Thus we have in the RHS to express $\displaystyle H_3''$, so we differentiate the second equation:

$\displaystyle H_3''= eH_2' - fH_3' +gH_4'$

We now substitute $\displaystyle H_2' \ and \ H_3'$ from the system equations and plug everything together:

$\displaystyle H_4'''= j( eH_2' - fH_3' +gH_4') - kH_4''$
$\displaystyle H_4'''= j( e[-dH_2- dH_3+ P]-f[eH_2 - fH_3 +gH_4]+gH_4') - kH_4''$

or:

(1) $\displaystyle H_4'''= -jedH_2-jedH_3+jeP-feH_2+f^2H_3+fgH_4+jgH_4'-kH_4''$

We'll now try express $\displaystyle H_2\ and\ H_3$ from the system:

$\displaystyle H_3=\frac{H_4'+kH_4}{j}$

$\displaystyle H_3'=\frac{H_4''+kH_4'}{j}=eH_2-\frac{f}{j}(H_4'+kH_4)+gH_4$

$\displaystyle H_4''+kH_4'=jeH_2-f(H_4'+kH_4)+jgH_4$

solving for H_2, we get:

(2)$\displaystyle H_2=\frac{H_4''+kH_4'+f(H_4'+kH_4)-jgH_4}{je}$

Now we resubstitute H_2 and H_3 in (1) to finish up with a 3rd order equation, where $\displaystyle cH_1=P$ and we get H_1 by solving the first equation of the system (it's linear and inhomogeneous)

Dealing with P

We take again the 2nd equation of the system:
$\displaystyle \dot H_2=-dH_2- dH_3+ P$
We want to solve for P and we already have H_2 and H_3 in terms of H_4, so we just differentiate (2)

$\displaystyle H_2'=\frac{H_4'''+kH_4''+f(H_4''+kH_4')-jgH_4'}{je}$

Solving for P we see the 1st equation of the system actually does not belong to it(cause the highest order of our DE still remains 3). It is considered as an inhomogeneous part of the 2nd equation of the system.

(I hope you haven't omitted something in this first equation )

That's all - only algebra

Best wishes, Marine

Thank a lot Marine, that really helped, sorry for the late reply

Cheers