106. Solve the equation $y^{\prime}\left(y^{2}-x\right)=y$.
Solution. This equation is not linear with respect to the unknown function $y$, since it cannot be brought to the form $y^{\prime} + p(x) y = q(x)$. However, if $x$ is considered a function and $y$-- an argument, then this equation will be linear with respect to the unknown function $x$, since it can be brought to the form $x^{\prime} + p(y) x = q(y)$.

Indeed, replacing $y^{\prime}$ with $\frac{d y}{d x}$, we get

$$
\frac{d y}{d x}\left(y^{2}-x\right)=y \text { or }\left(y^{2}-x\right) d y=y d x
$$

Dividing both sides of the last equation by the product $y d y$, we bring it to the form

$$
y-\frac{1}{y} x=x^{\prime} \text { or } x^{\prime}+\frac{1}{y} x=y
$$

Here $p(y)=\frac{1}{y}, q(y)=y$. This is a linear equation with respect to $x$; therefore, we assume $x=u v$ (where $u$ and $v$ are functions of $y$), from which $\frac{d x}{d y} = \frac{d u}{d y} v + \frac{d v}{d y} u$.

Substituting the expressions for $x$ and $\frac{d x}{d y}$ into equation (6), we have

$$
\frac{d u}{d y} v + \frac{d v}{d y} u + \frac{1}{y} u v = y \text { or } \frac{d u}{d y} v + \left(\frac{d v}{d y} + \frac{1}{y} v\right) u = y
$$

Choose $v$ such that

$$
\frac{d v}{d y} + \frac{1}{y} v = 0 \text { or } \frac{d v}{v} = -\frac{d y}{y}
$$

from which

$$
\int \frac{d v}{v} = -\int \frac{d y}{y} ; \int \frac{d v}{v} = -\int \frac{d y}{y} ; \ln |v| = -\ln |y| ; v = \frac{1}{y}
$$

Now equation (7) will take the form

$$
\frac{d u}{d y} \cdot \frac{1}{y} = y \text { or } d u = y^{2} d y
$$

from which

$$
\int d u = \int y^{2} d y + C, \text { i.e., } u = \frac{y^{3}}{3} + C
$$

Since $x = u v$, the general solution of the given equation is

$$
x = \left(\frac{y^{3}}{3} + C\right) \frac{1}{y}
$$