Example 2. Solve the equation $2\left(y^{\prime}\right)^{2}-2 x y^{\prime}-2 y+x^{2}=0$.
Solution. Let's solve the equation with respect to $y$:

$$
y=\left(y^{\prime}\right)^{2}-x y^{\prime}+\frac{x^{2}}{2}
$$

Let $y^{\prime}=p$, where $p$ is a parameter; then we get

$$
y=p^{2}-x p+\frac{x^{2}}{2}
$$

Differentiating (2), we find

$$
d y=2 p d p-p d x-x d p+x d x
$$

But since $d y=p d x$, we have

$$
p d x=2 p d p-p d x-x d p+x d x
$$

or

$$
\begin{aligned}
2 p d p-2 p d x-x d p+x d x & =0 \\
2 p(d p-d x)-x(d p-d x) & =0 \\
(2 p-x)(d p-d x) & =0
\end{aligned}
$$

Consider two cases:

1) $d p-d x=0$, from which $p=x+C$, where $C$ is an arbitrary constant. Substituting the value of $p$ into (2), we get the general solution of the given equation:

$$
y=C x+C^{2}+\frac{x^{2}}{2}
$$

In the equality $p=x+C$, we cannot replace $p$ with $y^{\prime}$ and integrate the resulting equation $y^{\prime}=x+C$ (since this would introduce a second arbitrary constant, which cannot be, as the differential equation under consideration is a first-order equation).
2) $2 p-x=0$, from which $p=\frac{x}{2}$. Substituting into (2), we get another solution

$$
y=\frac{x^{2}}{4}
$$

Let's check if the uniqueness property is violated at each point of the solution (4), i.e., whether it is a singular solution (see § 11). For this, take an arbitrary point $M_{0}\left(x_{0}, y_{0}\right)$ on the integral curve (4), where $y_{0}=\frac{x_{0}^{2}}{4}$. We will now find a solution that is contained in the general solution (3) and whose graph passes through the point $M_{0}\left(x_{0}, \frac{x_{0}^{2}}{4}\right)$. Substituting the coordinates of this point into the general solution (3), we get

$$
\frac{x_{0}^{2}}{4}=C x_{0}+C^{2}+\frac{x_{0}^{2}}{2}, \quad \text { or }\left(C+\frac{x_{0}}{2}\right)^{2}=0
$$

from which $C=-\frac{x_{0}}{2}$. Substituting this value of the constant $C$ into (3), we get the particular solution

$$
y=\frac{x^{2}}{2}-\frac{x_{0} x}{2}+\frac{x_{0}^{2}}{4}
$$

which does not coincide with the solution (4). For the solutions (4) and (5), we have $y^{\prime}=\frac{x}{2}, y^{\prime}=x-\frac{x_{0}}{2}$. At $x=x_{0}$, both derivatives coincide. Therefore, the uniqueness property is violated at the point $M_{0}$, i.e., through this point pass two integral curves with the same tangent. Since $x_{0}$ is arbitrary, the uniqueness is violated at each point of the solution (4), which means that it is a singular solution.

## Problems for Independent Solution

Integrate the following equations:

199. $4\left(y^{\prime}\right)^{2}-9 x=0$.
200. $\left(y^{\prime}\right)^{2}-2 y y^{\prime}=y^{2}\left(e^{2 x}-1\right)$.
201. $\left(y^{\prime}\right)^{2}-2 x y^{\prime}-8 x^{2}=0$.
202. $x^{2}\left(y^{\prime}\right)^{2}+3 x y y^{\prime}+2 y^{2}=0$.
203. $\left(y^{\prime}\right)^{2}-(2 x+y) y^{\prime}+x^{2}+x y=0$.
204. $\left(y^{\prime}\right)^{3}+(x+2) e^{y}=0$.
205. $\left(y^{\prime}\right)^{3}-y\left(y^{\prime}\right)^{2}-x^{2} y^{\prime}+x^{2} y=0$.
206. $\left(y^{\prime}\right)^{2}-4 x y^{\prime}+2 y+2 x^{2}=0$.
207. $\left(y^{\prime}\right)^{2}-y y^{\prime}+e^{x}=0$

$2^{\circ}$. Equations of the form $f\left(y, y^{\prime}\right)=0$ and $f\left(x, y^{\prime}\right)=0$. If the equations $f\left(y, y^{\prime}\right)=0$ and $f\left(x, y^{\prime}\right)=0$ are easily solvable with respect to $y^{\prime}$, then, solving them, we obtain equations with separable variables.

Consider the cases when these equations are not solvable with respect to $y^{\prime}$.

A. The equation of the form $f\left(y, y^{\prime}\right)=0$ is solvable with respect to $y$:

$$
y=\varphi\left(y^{\prime}\right)
$$

Let $y^{\prime}=p$, then $y=\varphi(p)$. Differentiating this equation and replacing $d y$ with $p d x$, we get

$$
p d x=\varphi^{\prime}(p) d p
$$

from which

$$
d x=\frac{\varphi^{\prime}(p)}{p} d p \quad \text { and } \quad x=\int \frac{\varphi^{\prime}(p)}{p} d p+C
$$

We obtain the general solution of the equation in parametric form

$$
x=\int \frac{\varphi^{\prime}(p)}{p} d p+C, \quad y=\varphi(p)
$$