Coupled differential equations

  • Thread starter plasmoid
  • Start date
  • #1
15
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I have the equations

[tex]\frac{l}{u^{2}} \frac{du}{dx}=constant[/tex]

and

[tex]\frac{1}{u} \frac{dl}{dx}=constant[/tex].

By "eyeball", I can say the solution is
[tex]l \propto x^{n} \ and \ u \propto x^{n-1}[/tex].

I can't see how I could arrive at these solutions 'properly', if you know what I mean
 

Answers and Replies

  • #2
82
2
Substract the first equation from the second, and observe that what you got on the left hand side is just the derivative of (l/u).
 
  • #3
AlephZero
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Alternatively, divide one equation by the other

[tex]\frac{l}{u} \frac{du}{dl} = c[/tex]

which has the general solution

[tex]u = l^p[/itex]

Substitute that in the second equation .....
 
  • #4
HallsofIvy
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Those equations are, in fact, "partially uncoupled"- the first equation, for u, does not depend on l. What I would do is just go ahead and solve the first equation for u, without regard for the second equation.
[tex]\frac{1}{u^2}\frac{du}{dx}= C[/tex]
so
[tex]\frac{du}{u^2}= Cdx[/tex]
integrating,
[tex]-\frac{1}{u}= Cx+ D[/tex]
so that
[itex]u= -\frac{1}{Cx+ D}[/tex]

Now put that into the second equation:
[tex]\frac{1}{u}\frac{dl}{dx}= -(Cx+ D}\frac{dl}{dx}= E[/tex]
[tex]-dl= -E(Cx+ D)dx[/tex]
so
[tex]-l(x)= -\frac{EC}{2}x^2- EDx+ F[/tex]
 
  • #5
15
0
Thanks guys ... I had started on the "divide one equation by the other" path, but for some reason did not carry it to it's conclusion.

@HallsofIvy, the first equation actually does depend on [tex]l[/tex]; I guess you mistook the [tex]l[/tex] in the numerator for [tex]1[/tex]. Thanks anyway :)
 
  • #6
HallsofIvy
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You are right about that- sorry.
 

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