Example 2. Find the general solution of the equation $\left(1+x^{2}\right) y^{\prime \prime}+y^{\prime 2}+$ $+1=0$ and the particular solution that satisfies the initial conditions $x=0, y=1, y^{\prime}=1$.
Solution. The given equation does not explicitly contain the variable $y$, so according to 2), we will use the substitution $y^{\prime}=p, y^{\prime \prime}=p^{\prime}$. This leads to a separable equation in terms of $p$: $\left(1+x^{2}\right) p^{\prime}+p^{2}+1=0, p^{\prime}+\frac{1+p^{2}}{1+x^{2}}=0$, i.e., $\frac{d p}{1+p^{2}}=\frac{-d x}{1+x^{2}}$. From this, $\operatorname{arctg} p=-\operatorname{arctg} x+\operatorname{arctg} C$, or $p=\operatorname{tg}(\operatorname{arctg} C-\operatorname{arctg} x)=\frac{C-x}{1+C x}$.

It remains to integrate the simple equation of the form 1) (nonlinear with separated variables) $y^{\prime}=\frac{C-x}{1+C x}$. We obtain
$y=-\frac{1}{C} \int \frac{\left(C x+1-C^{2}-1\right) d x}{C x+1}=-\frac{1}{C}\left(x-\frac{C^{2}+1}{C} \ln |C x+1|\right)+C_{1}-$ this is the general solution. To find the desired particular solution, we substitute the values $x=0, y=1, y^{\prime}=1$ into this and the expression for $y^{\prime}$ and determine $C=1, C_{1}=1$.

We substitute these values into the expression for $y$, from which we obtain the desired particular solution.

Answer. $y=-\frac{1}{C}\left(x-\frac{C^{2}+1}{C} \ln |C x+1|\right)+C_{1} ; \quad y_{4}=-x+$ $+2 \ln |x+1|+1$