4.19. Obtain the general solution of the equation $(x+y) d x+$ $+(x-y) d y=0$.
Solution. The functions $P(x, y)=x+y$ and $Q(x, y)=x-y$ are homogeneous functions of the first degree: $P(t x, t y)=$ $=t x+t y=t(x+y)=t P(x, y), Q(t x, t y)=t x-t y=t(x-y)=$ $=t Q(x, y)$. Introduce a new function $u(x)=\frac{y(x)}{x}$, hence $y(x)=x u(x), d y=u d x+x d u$. Substitute these into the original equation $(x+x u) d x+(x-x u)(u d x+x d u)=0$. Collect coefficients of the differentials: $\left(1+2 u-u^{2}\right) d x+$ $+x(1-u) d u=0$. Separate the variables: $\frac{(u-1) d u}{1+2 u-u^{2}}=\frac{d x}{x}$.

Integrate both sides, noting that $d\left(1+2 u-u^{2}\right)=$ $=(2-2 u) d u=-2(u-1) d u \quad$ and $\quad(u-1) d u=-\frac{1}{2} d\left(1+2 u-u^{2}\right)$, we have

$$
\int \frac{-0.5 d\left(1+2 u-u^{2}\right)}{1+2 u-u^{2}}=\int \frac{d x}{x},-\frac{1}{2} \ln \left|1+2 u-u^{2}\right|+\ln C=\ln |x|,
$$

from which $x \sqrt{1+2 u-u^{2}}=C$, returning to $y(x)=x u(x)$, we get $x^{2}+2 x y-y^{2}=C^{2}(C \neq 0)$.