Example 4. Solve the equation $y^{2 / 3}+\left(y^{\prime}\right)^{2 / 3}=1$.
Solution. Let $y=\cos ^{3} t, p=\sin ^{3} t$,

$$
d x=\frac{d y}{p}=\frac{-3 \cos ^{2} t \sin t d t}{\sin ^{3} t}=-3 \frac{\cos ^{2} t}{\sin ^{2} t} d t
$$

From this

$$
x=\int\left(3-\frac{3}{\sin ^{2} t}\right) d t=3 t+3 \operatorname{ctg} t+C
$$

the general solution is

$$
x=3 t+3 \operatorname{ctg} t+C, \quad y=\cos ^{3} t
$$

V. An equation of the form $f\left(x, y^{\prime}\right)=0$. Suppose this equation is solvable for $x$ :

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

Letting $y^{\prime} \equiv p$, we get $d x=\varphi^{\prime}(p) d p$. But $d x=\frac{d y}{p}$ and, therefore, $\frac{d y}{p}=\varphi^{\prime}(p) d p$, so

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

Thus

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

- the general solution of the equation in parametric form ( $p$ - parameter).

Remark. In formulas (6) and (7), $p$ cannot be considered as a derivative. In them, $p$ is simply a parameter.