Example 5. Solve the equation $a \frac{d y}{d x}+b\left(\frac{d y}{d x}\right)^{2}=x$.
Solution. Let $\frac{d y}{d x}=p$, then

$$
\begin{aligned}
x & =a p+b p^{2}, \quad d x=a d p+2 b p d p \\
d y & =p d x=a p d p+2 b p^{2} d p, \quad y=\frac{a}{2} p^{2}+\frac{2}{3} b p^{3}+C
\end{aligned}
$$

Thus,

$$
x=a p+b p^{2}, \quad y=\frac{a}{2} p^{2}+\frac{2}{3} b p^{3}+C-\text { general solution. }
$$

Similarly to case B, one can try to solve the equation $f\left(x, y^{\prime}\right)=0$ by introducing a parameter $t$.

## Problems for Independent Solution

Integrate the following equations:

208. $y=\left(y^{\prime}\right)^{2} e^{y^{\prime}}$
209. $x=\ln y^{\prime}+\sin y^{\prime}$.
210. $y=y^{\prime} \ln y^{\prime}$.
211. $\left(y^{\prime}\right)^{2} x=e^{1 / y^{\prime}}$.
212. $y^{2 / 5}+\left(y^{\prime}\right)^{2 / 5}=a^{2 / 5}$.
213. $y=y^{\prime}\left(1+y^{\prime} \cos y^{\prime}\right)$.
214. $y^{\prime}=e^{y^{\prime} / y}$
215. $x=\left(y^{\prime}\right)^{2}-2 y^{\prime}+2$.
216. $y=\left(y^{\prime}-1\right) e^{y^{\prime}}$.
217. $x\left(1+\left(y^{\prime}\right)^{2}\right)^{3 / 2}=a$.
218. $x=y^{\prime}+\sin y^{\prime}$.
219. $y=\arcsin y^{\prime}+\ln \left(1+\left(y^{\prime}\right)^{2}\right)$.

$3^{\circ}$. Equations of Lagrange and Clairaut. The Lagrange equation has the form

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

By setting $y^{\prime}=p$, differentiating with respect to $x$, and replacing $d y$ with $p d x$, we reduce this equation to a linear equation in $x$ as a function of $p$. Finding the solution of this last equation $x=r(p, C)$, we obtain the general solution of the original equation in parametric form:

$$
x=r(p, C), \quad y=r(p, C) \varphi(p)+\psi(p) \quad(p-\text { parameter }) .
$$

In addition, the Lagrange equation may have special solutions (see §11) of the form $y=\varphi(c) x+\psi(c)$, where $c$ is a root of the equation $c=\varphi(c)$.