Differential equation with $\sqrt{xy}$ $$7\sqrt{xy} \frac{dy}{dx}=4, \quad x,y>0$$
How do I solve this equation for $y$
 A: Hint: The equation is separable and
$$ \sqrt[7]{xy} = \sqrt[7]{x} \sqrt[7]{y} $$
A: If an equation is invariant to $x'=\lambda x$ and $y'=\lambda^\beta y$ then $$\frac{dy}{dx}=\beta\frac{y}{x}$$ $$\frac{d^2y}{dx^2}=\beta(\beta -1)\frac{y}{x^2}$$ $$\frac{d^3y}{dx^3}=\beta(\beta -1)(\beta -2)\frac{y}{x^3}$$ and so on.  These are Lie Autodiffeomorphisms, and not too many people know about them, but they work dandy for problems like yours. They don't provide complete solutions, just special solutions without integration constants (because they don't require integration to use).  
Just apply the group transformation to your ODE:
$$
7\sqrt{\lambda x\lambda^\beta y}\frac{d\lambda^\beta y}{d\lambda x}=4
$$ 
$$
\large\lambda^{\frac{1}{2}+\frac{\beta}{2}+\beta -1}7\sqrt{xy}\frac{dy}{dx}=4
$$
For invariance,
$$
\large\lambda^{\frac{3\beta}{2}-\frac{1}{2}}=1
$$so $\beta=\frac{1}{3}$  Therefore, 
$$
\frac{dy}{dx}=\frac{y}{3x}
$$Make this substitution and do a little algebra, et voila!
$$
\large y=\bigg(\frac{12}{7}\bigg)^\frac{2}{3}x^\frac{1}{3}
$$
Separation of variables provides the general solution:
$$
y=\bigg(\frac{12}{7}x^\frac{1}{2}+C\bigg)^\frac{2}{3}
$$Set $C=0$ and this collapses to the special solution, as expected.
NOTE:  If your original problem was 
$$
(xy)^\frac{1}{7}\frac{dy}{dx}=4
$$this just means $\beta=\frac{3}{4}$ and the special solution is 
$$
y=\bigg(\frac{16}{3}\bigg)^\frac{7}{8}x^\frac{3}{4}
$$Not sure which one you asked, but the principle is the same either way.
A: Have you tried separating the differential equation? You should get 
$$y^{1/7} \, dy = 4x^{-1/7} \, dx.$$
Now you need to integrate both sides of that equation.
