Example 6. Solve the Clairaut equation $y^{\prime}=x y^{\prime}+\frac{3}{y^{\prime 2}}$.
Solution. The Clairaut equation $y=x y^{\prime}+\psi\left(y^{\prime}\right)$ (here $\psi\left(y^{\prime}\right)=\frac{3}{y^{\prime 2}}$) is solved in two stages.

1) Notice that the substitution $y^{\prime}=C(C \neq 0-$ an arbitrary constant) satisfies the given equation: from it follows that $y=x \cdot C+C_{1} ;$ setting $C_{1}=\frac{3}{C^{2}}$ here, we arrive at the identity $C x+\frac{3}{C^{2}} \equiv C x+\frac{3}{C^{2}}$ (the left side is obtained by integrating the equality $y^{\prime}=C$, and the right side is taken from the equation with the substitution $y^{\prime}=C$). Therefore, the equality $y=C x+\frac{3}{C^{2}}$ gives the general solution of the original equation.
2) Finding the general solution. Let us set in the original equation, as usual, $y^{\prime}=t$, and then differentiate the obtained equality with respect to $x$ with the subsequent replacement of $y^{\prime}$ by $t$:

$$
y=x t+\frac{3}{t^{2}}, \quad y^{\prime}=t+t^{\prime} x-\frac{6 t^{\prime}}{t^{3}} \Leftrightarrow t=t+t^{\prime} x-\frac{6 t^{\prime}}{t^{3}}, t^{\prime}\left(x-\frac{6}{t^{3}}\right)=0
$$

We analyze this equality:

a) $t^{\prime}=0 \Leftrightarrow t=C \Leftrightarrow y^{\prime}=C$; we obtain the relation already considered in point 1); it leads to the general solution;

b) $x-\frac{6}{t^{3}}=0, \quad x=\frac{6}{t^{3}}$ (dependence of $x$ on the parameter $t$).

Adding this expression to the dependence of $y$ on $t$, we obtain the parametric solution: $x=\frac{6}{t^{3}}, \quad y=x t+\frac{3}{t^{2}}$.

Incidentally, the given form of the function $\psi\left(y^{\prime}\right)$ allows us to explicitly express the parameter $t$ in terms of $x$ and thereby obtain the explicit function $y=y(2)$ - the solution of the original equation. Let's do this: $x=\frac{6}{t^{3}} \Leftrightarrow t=\sqrt[3]{\frac{6}{x}}$; substituting into the equality $y=x t+\frac{3}{t^{2}}$, we get $y=x \cdot \sqrt[3]{\frac{6}{x}}+3 \sqrt[3]{\frac{x^{2}}{36}}=$ $=\frac{\sqrt[3]{x^{2}}}{\sqrt[3]{4}}(\sqrt[3]{24}+\sqrt[3]{3})=\frac{3 \sqrt[3]{3 x^{2}}}{\sqrt[3]{4}}$. Hence $y^{3}=\frac{81}{4} x^{2}$ (the simplest implicit form of the solution). This is the so-called special (not particular) solution of the original equation, which cannot be obtained from the general solution for any value of the constant $C$.

$$
\text { Answer. } y_{\text {gen }}=C x+\frac{3}{C^{2}}, \quad y_{\text {spec}}=\sqrt[3]{\frac{81}{4} x^{2}}
$$

Note. There is a connection between the general and special solutions of Clairaut equations, which allows for a geometric illustration.
The graph $\Gamma$ of the special solution represents a semi-cubic parabola $y^{3}=\frac{81}{4} x^{2}$, symmetric with respect to the $y$-axis (Fig. 4.3). It turns out that the general solution is a family of lines $y=C x+\frac{3}{C^{2}}$ ( $C$ - an arbitrary constant,

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-090.jpg?height=284&width=524&top_left_y=185&top_left_x=1961)

Fig. 4.3 $C \neq 0)$, and each line of the family is tangent to $\Gamma$, while the point $(x, y)$ on the graph of $\Gamma$ belongs simultaneously to some line $y=C x+\frac{3}{C^{2}}$. It is said that the special solution forms the envelope of the family of lines (general integral curves).

In general, suppose that some differential equation has a general solution with the following property: there exists an integral curve that does not belong to this family but has a common point with each integral curve of the general solution. Then this new line is the envelope of the first family and a special solution of the original equation. This is exactly the line $y^{3}=\frac{81}{4} x^{2}$ for the equation in Example 6.