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

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

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
[itex]u= -\frac{1}{Cx+ D}[/tex]

Now put that into the second equation:
$$\frac{1}{u}\frac{dl}{dx}= -(Cx+ D}\frac{dl}{dx}= E$$
$$-dl= -E(Cx+ D)dx$$
so
$$-l(x)= -\frac{EC}{2}x^2- EDx+ F$$

 

[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.