How Do You Solve These Coupled Differential Equations?

[plasmoid]

I have the equations

\frac{l}{u^{2}} \frac{du}{dx}=constant

and

\frac{1}{u} \frac{dl}{dx}=constant.

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

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

 

[csopi]

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).

 

[AlephZero]

Alternatively, divide one equation by the other

\frac{l}{u} \frac{du}{dl} = c

which has the general solution

u = l^p[/itex]<br /> <br /> Substitute that in the second equation ...

 

[HallsofIvy]

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.
\frac{1}{u^2}\frac{du}{dx}= C
so
\frac{du}{u^2}= Cdx
integrating,
-\frac{1}{u}= Cx+ D
so that
u= -\frac{1}{Cx+ D}[/tex]<br /> <br /> Now put that into the second equation:<br /> \frac{1}{u}\frac{dl}{dx}= -(Cx+ D}\frac{dl}{dx}= E<br /> -dl= -E(Cx+ D)dx<br /> so<br /> -l(x)= -\frac{EC}{2}x^2- EDx+ F

 

[plasmoid]

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 l; I guess you mistook the l in the numerator for 1. Thanks anyway :)

 

[HallsofIvy]

You are right about that- sorry.