Solving the differential equation $7x(x-y)dy = 2(x^2+6xy-5y^2)dx$ How do I solve the differential equation $$7x(x-y)dy = 2(x^2+6xy-5y^2)dx$$
Is it homogeneous? I have tried taking the variables from the LHS and applying them to the RHS, making $\frac{dy}{dx}$ subject and ending up with:
$$\frac{dy}{dx} = \frac{2(x^2+6xy-5y^2)}{7x(x-y)}$$
After simplifying the numerator: 
$$\frac{dy}{dx} = \frac{2[(x-y)(x+5y)+2xy]}{7x(x-y)}$$
$$\frac{dy}{dx} = \frac{2(x+5y)}{7x} + \frac{4y}{7(x-y)}$$
I have no idea how to proceed from here. Is the methodology correct so far? Any input would be appreciated.
 A: Your differential equation is indeed homogeneous.
Therefore, you can substitute $y=vx$ and $\frac{dy}{dx}=x\frac{dv}{dx}+v$, where (NB: $v$ is a function of $x$). The substitution for $\frac{dy}{dx}$ is evaluated using implicit differentiation.
$$\frac{dy}{dx}=\frac{2(x^2+6xy-5y^2)}{7x(x-y)}$$
$$x\frac{dv}{dx}+v=\frac{2(x^2+6x^2v-5v^2 x^2)}{7x(x-vx)}$$
$$x\frac{dv}{dx}+v=\frac{2(1+6v-5v^2)}{7(1-v)}$$
$$x\frac{dv}{dx}=\frac{2(1+6v-5v^2)}{7(1-v)}-v$$
$$x\frac{dv}{dx}=\frac{2(1+6v-5v^2)}{7(1-v)}-\frac{7v(1-v)}{7(1-v)}$$
$$x\frac{dv}{dx}=\frac{2(1+6v-5v^2)-7v(1-v)}{7(1-v)}$$
$$x\frac{dv}{dx}=\frac{2+12v-10v^2-7v+7v^2}{7(1-v)}$$
$$x\frac{dv}{dx}=\frac{2+5v-3v^2}{7(1-v)}$$
$$x\frac{dv}{dx}=\frac{3v^2-5v-2}{7(v-1)} \tag{1}$$
You will easily notice that equation $(1)$ is separable.
Now, you can rearrange equation $(1)$, and then integrate both sides.
$$\int \frac{7(v-1)}{3v^2-5v-2}~dv=\int \frac{1}{x}~dx$$
All you have to do now, is integrate both sides, and substitute back for $v=\frac{y}{x}$.
$$\ln|2-v|+\frac{4}{3} \ln|3v+1|=\ln|x|+C$$
$$\ln\left|2-\frac{y}{x}\right|+\frac{4}{3} \ln\left|\frac{3y}{x}+1\right|=\ln|x|+C \tag{2}$$
Thus, an implicit solution to your differential equation is on equation $(2)$. I am unsure whether an explicit solution for $y(x)$ exists, however if we check on Wolfram Alpha, we get a rather strange set of explicit solutions.
A: Starting with your equation
$$\frac{dy}{dx} = \frac{2(x+5y)}{7x} + \frac{4y}{7(x-y)}$$
multiply the top and bottom of each fraction on the right, giving
$$\frac{dy}{dx} = \frac{2(x+5y)}{7x}\frac{1/x}{1/x} + \frac{4y}{7(x-y)}\frac{1/x}{1/x}$$
$$\frac{dy}{dx} = \frac{2(1+5y/x)}{7} + \frac{4y/x}{7(1-y/x)}$$
Since $y'=u'x+u$ you get
$$u'x+u = \frac{2(1+5u)}{7} + \frac{4u}{7(1-u)}$$
which is a separable differential equation.
