Difficult Homogeneous Differential Equation Solve the differential equation:
$$\frac{dy}{dx}=\frac{\sqrt{x^2+3xy+4y^2}}{x+2y}$$
I tried to solve it by putting $t=x+2y$ but that lead to a very complicated integral. The hint given is that equation is reducible to homogeneous form.

Partial answer :
Reformulate it as $$\frac{dy}{dx} = \frac{\sqrt{1+3\frac{y}{x}+4\left(\frac{y}{x}\right)^2}}{1+2\frac{y}{x}}$$
The you have $dy/dx = f(y/x)$.
Let $u(x) = \frac{y(x)}{x}$. Then $xu = y$. Then $dy = xdu + udx$. Then $dy/dx = xdu/dx + u$. Then you have
$$x\frac{du}{dx}+u = f(u)\quad \text{where} \quad f(u) = \frac{\sqrt{1+3u+4u^2}}{1+2u}$$
Then $x\frac{du}{dx} = f(u)-u$. Then
$$\int \frac{du}{f(u)-u} = \int\frac{dx}{x}$$
i.e.
$$\int \frac{1+2u}{\sqrt{4u^2+3u+1}-u-2u^2} = \int\frac{dx}{x}$$
So this must be solved (to get further from here). It'd give you $u(x)$ as a function of $x$ and from there you get $y(x)$ using $y(x)=x u(x)$. Here's a beginning :
$$\frac{1+2u}{\sqrt{4u^2+3u+1}-(u+2u^2)} = \frac{(1+2u)(\sqrt{4u^2+3u+1}+u+2u^2)}{4u^2+3u+1-(u+2u^2)^2}
\\ = -\frac{(1+2u)(\sqrt{4u^2+3u+1}+u+2u^2)}{4u^4 + 4u^3 - 3u^2 - 3u -1}
\\ = -\frac{(\sqrt{4u^2+3u+1}+u+2u^2)}{2u^3 + 2u^2 - 2u - \frac{u+1}{2u+1}}
\\ = \text{ugly...}$$
Then : either goto Claude Leibovici's highly plausible answer or please post yours if you know how to solve this.
