Solve the differential equation $(x + y)\,\mathrm{d}y = (4x + y)\,\mathrm{d}x$ 
$$(x + y)\,\mathrm{d}y = (4x + y)\,\mathrm{d}x$$  

The variables don't seem to be separable, so I divided both sides by $x$ and put $ t = \frac y x$. Even after this, I get a non-separable equation.
 A: It is separable: If we let $y=tx$, we get that
$$\frac{\mathrm{d}y}{\mathrm{d}x}=x\frac{\mathrm{d}t}{\mathrm{d}x}+t$$
So
$$\frac{4x+tx}{x+tx}=x\frac{\mathrm{d}t}{\mathrm{d}x}+t$$
$$\frac{4+t}{1+t}=x\frac{\mathrm{d}t}{\mathrm{d}x}+t$$
$$\frac{4+t}{1+t}-t=x\frac{\mathrm{d}t}{\mathrm{d}x}$$
$$\frac{4+t}{1+t}-t\frac{1+t}{1+t}=x\frac{\mathrm{d}t}{\mathrm{d}x}$$
$$\frac{4+t}{1+t}-\frac{t+t^2}{1+t}=x\frac{\mathrm{d}t}{\mathrm{d}x}$$
$$\frac{4-t^2}{1+t}=x\frac{\mathrm{d}t}{\mathrm{d}x}$$
$$\frac{1+t}{4-t^2}=\frac{1}{x}\frac{\mathrm{d}x}{\mathrm{d}t}$$
A: This equation is homogeneous of order $0$ and we write
$$y'=\dfrac{4x+y}{x+y}=\dfrac{4+\frac{y}{x}}{1+\frac{y}{x}}$$
Let $u=\dfrac{y}{x}$ then $u'x+u=y$ with substitution
$$u'x+u=\dfrac{4+u}{1+u}$$
or
$$\dfrac{1+u}{4-u^2}du=\dfrac{dx}{x}$$
implies 
$$-\dfrac14\ln(u+2)-\dfrac34\ln(2-u)=\ln(x)+C$$
now we replace $u=\dfrac{y}{x}$.
A: HINT: Note that $$\frac{\partial}{\partial x} (x + y) = 1 = \frac{\partial}{\partial y} (4x + y) $$ and think of the solution in the form of $F(x, y) = 0$.
Can you proceed from here?
EDIT: As it was correctly noticed by Isham I made a mistake and the ODE $$(x + y) dy = (4x + y) dx$$
is indeed not exact. Below is the method that I was initially talking about.

In general, we have 
$$M(x, y) dx + N(x, y) dy = 0,$$
which is said to exact if $M_y - N_x = 0$, where the subscript variable denotes the argument of differentiation. It is straightforward to obtain a solution from here in the form $F(x, y) = 0$. For details see here.
However, in some cases when $M_y \neq N_x$ it still can be made to be exact. The latter is called Inexact differential equation. Unfortunately, here it is not the case, since both 
$$\frac{M_y - N_x}{M} = \frac 2 {x+y}$$
and 
$$\frac{M_y - N_x}{N} = \frac 2 {4x + y}$$
depends on both $x$ and $y$ and hence the initial ODE is not inexact either. 
A: $$(x + y)\,\mathrm{d}y = (4x + y)\,\mathrm{d}x$$
$$\frac {dx}{(x + y)} = \frac {dy}{(4x + y)}$$
$$\frac {dx}{(x + y)} = \frac {dy}{(4x + y)}=-\frac {d(y-2x)}{(y-2x)} = \frac {d(2x+y)}{3(2x + y)}$$
We work with these 2 last fractions
$$-\frac {d(y-2x)}{(y-2x)} = \frac {d(2x+y)}{3(2x + y)}$$
After integration 
$$-3\ln(y-2x) = \ln(2x + y)+K$$
Finally we get :
$$\boxed {(y-2x)^3(y+2x)=C}$$
