33. Solve the equation $\frac{d y}{y+1}=\frac{d x}{x-1}$.
Solution. Here the variables are separated. Integrating, we have

$$
\int \frac{d y}{y+1}=\int \frac{d x}{x-1} ; \ln (y+1)=\ln (x-1)+C
$$

The arbitrary constant $C$ can be denoted as $\ln C$; then $\ln (y+1)=\ln (x-1)+\ln C$. Representing the sum of logarithms on the right side of the equation as the logarithm of a product, we get $\ln (y+1)=$ $=\ln C(x-1)$, from which $y+1=C(x-1)$. This is the general integral of the given differential equation.

From a geometric point of view, we have obtained the equation of a pencil of lines centered at the point $M(1, -1)$ with a slope $C$ (compare the obtained equation with the equation of a pencil of lines $y-y_{1}=k(x-$ $\left.-x_{1}\right)$ ).

34-39. Solve the equations: