4.21. Obtain the general solution of the equation

$$
\frac{d y}{d x}=\frac{x+y+1}{4 x+4 y-1}
$$
Solution. Let's form the determinant $\Delta=\left|\begin{array}{ll}1 & 1 \\ 4 & 4\end{array}\right|=0$. We have the third case. Introduce the substitution $z=x+y, 4 x+4 y=4 z, d y=d z-$ $-d x, \frac{d y}{d x}=\frac{d z}{d x}-1$. The equation takes the form $\frac{d z}{d x}-1=\frac{z+1}{4 z-1}$. Separating variables: $\frac{(4 z-1) d z}{5 z}=d x$. Integrating:

$$
\frac{4}{5} \int d z-\frac{1}{5} \int \frac{d z}{z}=x+C, \frac{4}{5} z-\frac{1}{5} \ln |z|=x+C
$$

Returning to the original variables:

$$
\frac{4}{5}(x+y)-\frac{1}{5} \ln |x+y|=x+C
$$