Solving coupled ODEs Context: This is part of one of the derivation I'm trying to do for turbulent wake flows. The book contains the governing equations and the answer but with my limited mathematical knowledge, I'm not able to see through how to approach the problem.
Problem: We have two variables, $l(x)$ and $U(x)$ and they are related through these coupled ODEs:
$$\frac{C_o}{U} \frac{dl}{dx} = C_1$$
$$-\frac{lC_o}{U^2} \frac{dU}{dx} = C_2$$
Here, all the $C_i$ are constant and the final expression is given as $l = Ax^n$ and $U = Bx^{n-1}$ (again, A and B are constants). I can verify that these are indeed the correct answers (therefore, no typographical error).
My try: I did try to attempt this question in a way similar to a previous question I had asked on the site but to no luck. Perhaps, I'm looking for some nudge in the right direction and thereafter, I could do the rest of the calculations.
 A: You have two equations
$$
\frac{C_o}{U} \frac{dl}{dx} = C_1\\
lC_o \frac{dU}{dx} =lC_o\frac{d}{dx}\frac{1}{U} = C_2
$$
Let $v = 1/U$
we have
$$
C_o v \frac{dl}{dx} = C_1\\
C_o l\frac{dv}{dx} = C_2
$$
we can add the two together
$$
C_0 \left[v \frac{dl}{dx} + l\frac{dv}{dx}\right] = C_1 + C_2
$$
or
$$
C_0\frac{d}{dx}lv = C_1 + C_2
$$
so
$$
lv = C_3 x + C_4
$$
where $C_3 = \frac{C_1+C_2}{C_0}$
or
$$
\frac{l}{U} = C_3 x + C_4
$$
if $C_4 = 0$ then you can choose any $l$ and $U$ such that the quotient is a difference of $1$. However, this is usually defined by conditions.
A: Dividing the two we get $-\frac Ul\frac{dl}{dU}=\frac{C_1}{C_2}$ which is separable and upon integration yields$$\begin{align*}&-\int\frac{dl}l=\frac{C_1}{C_2}\int\frac{dU}U\\&\implies\log l=-\frac{C_1}{C_2}\log U+k\\&\implies l=k_1U^{-C_1/C_2}(*)
\end{align*}$$
Now substitute for $l$ in the second differential equation to get$$\frac{dU}{dx}=k_2U^{2+C_1/C_2}$$giving $U^{-C_1/C_2-1}=Ax+B$. Substitute back $U$ in $(*)$ to get $l$.
