4.3. Solve the Cauchy problem: $\frac{d y}{d x}=\frac{1}{(x+y)^{2}}, y(0)=1$.
The solution is as follows. $1 \cdot \check{i} s t e p$. Introduce a new unknown function of substitution $t=x+y$, then $\frac{d t}{d x}=1+\frac{1}{t^{2}}$.

$2-\check{u}$ step. Separate the variables: $\frac{t^{2} d t}{1+t^{2}}=d x$, integrate both sides, each with respect to its own variable:

$$
\int \frac{t^{2} d t}{1+t^{2}}=\int d x, \quad \int \frac{\left(t^{2}+1-1\right) d t}{1+t^{2}}=x+C, \int\left(1-\frac{1}{1+t^{2}}\right) d t=x+C
$$

$t-\operatorname{arctg} t=x+C,(x+y)-\operatorname{arctg}(x+y)=x+C$. We obtained the general solution in implicit form.

3rd step. Find the particular solution using the initial condition: $(0+1)-\operatorname{arctg}(0+1)=0+C, C=1-\frac{\pi}{4}$. The solution to the Cauchy problem is also written in implicit form: $(x+y)-\operatorname{arctg}(x+y)=x+1-\frac{\pi}{4}$.

4.2 .1 .2 PROBLEMS FOR INDEPENDENT SOLUTION

Integrate the equations.