43. Find the general solution of the equation $1+y^{\prime}+y+x y^{\prime}=0$.
Solution. $1^{0}$. Note $y^{\prime}$ as $\frac{d y}{d x}$ :

$$
1+\frac{d y}{d x}+y+x \frac{d y}{d x}=0
$$

$2^{0}$. Multiply all terms of the equation by $d x$ :

$$
d x+d y+y d x+x d y=0
$$

Group all terms containing $d y$ and $d x$, and write the obtained expressions in different parts of the equation:

$$
(1+x) d y=-(1+y) d x
$$

$3^{0}$. Divide both sides of the equation by the expression $(1+x)(1+y)$.

$$
\frac{d y}{1+y}=-\frac{d x}{1+x}
$$

$4^{0}$. Integrating both sides of the equation, we have

$$
\begin{aligned}
& \int \frac{d y}{1+y}=-\int \frac{d x}{1+x} ; \ln |1+y|=-\ln |1+x|+\ln C \\
& \ln |1+y|=\ln \left|\frac{C}{1+x}\right| ; 1+y=\frac{C}{1+x} ; y=\frac{C}{1+x}-1
\end{aligned}
$$

44-51. Solve the equations: