Example 4. $y=3\left(y^{\prime}\right)^{2}-5\left(y^{\prime}\right)^{3}$.


Solution. The given equation can be classified as a general form equation $\varphi\left(y, y^{\prime}\right)=0$, and we can apply the reasoning from section 6.2.

Let $y^{\prime}=t\left(t-\right.$ be a parameter). We obtain the equality $y=3 t^{2}-5 t^{3}$, which we differentiate with respect to $x$ (note that $t=y^{\prime}-$ is a function of $x, t=t(x)$, and then $\left.t^{\prime}=\frac{d t}{d x}=\left(\frac{d x}{d t}\right)^{-1}\right)$. We get $y^{\prime}=6 t \cdot t^{\prime}-15 t^{2} \cdot t^{\prime}$, or (replacing $y^{\prime}=t$) $t=3 t \cdot(2-5 t) \cdot t^{\prime}$. We analyze this equality.

1) If $t=0$, i.e., $y^{\prime}=0$, then the function $y=C$ is a solution to the obtained equation, but generally does not satisfy the original equation: $C=3\left(C^{\prime}\right)^{2}-5\left(C^{\prime}\right)^{3}$ holds only when $C=0$, which gives one trivial solution to the original equation: $y=0$.
2) If $t \neq 0$, we can represent the analyzed equation as $1=3(2-5 t) \cdot \frac{d t}{d x}$, or $d x=3(2-5 t) d t$. Its solution is $x=-\frac{3}{10}(2-5 t)^{2}+C_{1}$. This equality cannot define the general solution of the original equation, because $t$ is not an independent variable (since $t=y^{\prime}$) and, additionally, the obtained equality does not contain the variable $y$, and therefore there is no analytical expression of the integral curve, i.e., the dependence between $x$ and $y$.

We append the obtained equality to the original equation (equality) with $t=y^{\prime}$ as a parameter. The resulting system represents the general solution of the original differential equation in parametric form.

$$
\text { Answer. } .\left\{\begin{array}{l}
x=-\frac{3}{10}(2-5 t)^{2}+C_{\mathrm{I}}, \\
y=3 t^{2}-5 t^{3}
\end{array} \quad y=0\right.
$$