The solution of the differential equation $$\frac{{dy}}{{dx}} = \frac{{1 - 3y - 3x}}{{1 + x + y}}$$     is :

Solution :
This is the form in which $$\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}}$$
The given equation can be rewritten as $$\frac{{dy}}{{dx}} = \frac{{1 - 3\left( {x + y} \right)}}{{1 + \left( {x + y} \right)}} = f\left( {x + y} \right)$$
Substitute $$x + y = z \Rightarrow 1 + \frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}$$
The equation then becomes
$$\eqalign{ & \frac{{dz}}{{dx}} - 1 = \frac{{1 - 3z}}{{1 + z}} \cr & \Rightarrow \frac{{dz}}{{dx}} = \frac{{1 - 3z + 1 + z}}{{1 + z}} \cr & \Rightarrow \frac{{dz}}{{dx}} = \frac{{2 - 2z}}{{1 + z}} \cr & \Rightarrow \frac{{1 + z}}{{2\left( {1 - z} \right)}}dz = dx \cr} $$
On integrating we get
$$\eqalign{ & \frac{1}{2}\int {\frac{{1 + z}}{{1 - z}}dz = } \int {dx + a} \cr & \Rightarrow \frac{1}{2}\int {\left[ {\frac{2}{{1 - z}} - 1} \right]dz = x + a} \cr & \Rightarrow - \ell n\left| {1 - z} \right| - \frac{1}{2}z = x + a \cr & \Rightarrow - \ell n\left| {1 - x - y} \right| - \frac{1}{2}\left( {x + y} \right) = x + a \cr & \Rightarrow - 2\ell n\left| {1 - x - y} \right| - 3x - y = 2a \cr & \Rightarrow 3x + y + 2\ell n\left| {1 - x - y} \right| = c{\text{ where }}c = - 2a \cr} $$