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Math Help - Differential Equation Systems

  1. #1
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    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
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  2. #2
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    Hi there!

    All you do is just algebra

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

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

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

    rewrite the system using differentials:

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

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

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

    or

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

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

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

    ***

    But I suspect you meant something different:

    Let's take our system again:

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

    Now, we take the first equation and:

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

    (2). Express y with x: 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:

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

    and

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

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

    let's rearrange a bit:

    \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
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  3. #3
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    now, as I look at it once again I noticed following:

    Our system can be written as follows:

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

    The left hand side:

    \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:

    (a+d) is spur(A) and (ad-bc) is 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.
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  4. #4
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    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 = ?
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  5. #5
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    Hi there!

    Let's latex the system:

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

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

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

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

    H_4'''=           jH_3'' - kH_4''

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

    H_3''=  eH_2' - fH_3' +gH_4'

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

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

    or:

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

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

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

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

    H_4''+kH_4'=jeH_2-f(H_4'+kH_4)+jgH_4

    solving for H_2, we get:

    (2) 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 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:
    \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)

    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
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  6. #6
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    Differential Equations

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

    Cheers
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