Example 7. Integrate the equation $y=x y^{\prime}+\frac{a}{2 y^{\prime}}$ ( $a=$ const).
Solution. Setting $y^{\prime}=p$, we get

$$
y=x p+\frac{a}{2 p}
$$

Differentiating the last equation and replacing $d y$ with $p d x$, we find

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

from which

$$
d p\left(x-\frac{a}{2 p^{2}}\right)=0
$$

Setting the first factor to zero, we get $d p=0$, hence $p=C$ and the general solution of the original equation is $y=C x+\frac{a}{2 C}$, a one-parameter family of straight lines. Setting the second factor to zero, we have

![](https://cdn.mathpix.com/cropped/2024_05_22_a0612bb33256580e6dcdg-050.jpg?height=528&width=423&top_left_y=753&top_left_x=829)

Fig. 14

$$
x=\frac{a}{2 p^{2}}
$$

Eliminating $p$ from this equation and from the equation $y=x p+\frac{a}{2 p}$, we get $y^{2}=2 a x$ - this is also a solution of our equation (a particular solution).

From a geometric point of view, the curve $y^{2}=2 a x$ is the envelope of the family of straight lines given by the general solution (Fig. 14).

## Problems for Independent Solution

Integrate the following equations:

220. $y=2 x y^{\prime}+\ln y^{\prime}$.
221. $y=x\left(1+y^{\prime}\right)+\left(y^{\prime}\right)^{2}$.
222. $y=2 x y^{\prime}+\sin ^{2} y^{\prime}$.
223. $y=x\left(y^{\prime}\right)^{2}-\frac{1}{y^{\prime}}$.
224. $y=\frac{3}{2} x y^{\prime}+e^{y^{\prime}}$.
225. $y=x y^{\prime}+\left(y^{\prime}\right)^{2}$.
226. $y=x y^{\prime}+a \sqrt{1+\left(y^{\prime}\right)^{2}}$.
227. $y=x y^{\prime}+\frac{a}{\left(y^{\prime}\right)^{2}}$.
228. $x\left(y^{\prime}\right)^{2}-y y^{\prime}-y^{\prime}+1=0$.
229. $x=\frac{y}{y^{\prime}}+\frac{1}{\left(y^{\prime}\right)^{2}}$.
230. Find the curve such that each tangent to it forms a triangle of constant area $S=2 a^{2}$ with the coordinate axes.
231. Find the curve for which the segment of the tangent, enclosed between the coordinate axes, has a constant length $a$.

## § 9. Riccati Equation

A first-order differential equation of the form

$$
\frac{d y}{d x}+a(x) y^{2}+b(x) y+c(x)=0
$$

where $a(x), b(x), c(x)$ are known functions, is called a Riccati equation (generalized). If the coefficients $a, b, c$ in the Riccati equation are constant, the equation allows for separation of variables, and we immediately obtain the general integral

$$
C_{1}-x=\int \frac{d y}{a y^{2}+b y+c} .
$$

As Liouville showed, the equation (1) in general cannot be integrated in quadratures.

Properties of the Riccati equation. 1. If any particular solution $y_{1}(x)$ of equation (1) is known, then its general solution can be obtained using quadratures.

Indeed, let

$$
y=y_{1}(x)+z(x)
$$

where $z(x)$ is a new unknown function. Substituting (2) into (1), we get

$$
\frac{d y_{1}}{d x}+\frac{d z}{d x}+a(x)\left(y_{1}^{2}+2 y_{1} z+z^{2}\right)+b(x)\left(y_{1}+z\right)+c(x)=0
$$

from which, since $y_{1}(x)$ is a solution of equation (1), we obtain

$$
\frac{d z}{d x}+a(x)\left(2 y_{1} z+z^{2}\right)+b(x) z=0
$$

or

$$
\frac{d z}{d x}+a(x) z^{2}+\left[2 a(x) y_{1}+b(x)\right] z=0
$$

Equation (3) is a special case of the Bernoulli equation (see § 6).