Example 8. Solve the Cauchy problem $y^{\prime \prime}=2 y^{3} ;\left.y\right|_{x=0}=1,\left.y^{\prime}\right|_{x=0}=1$.
Solution. Setting $y^{\prime}=p$, we obtain

$$
p \frac{d p}{d y}=2 y^{3}
$$

from which

$$
p^{2}=y^{4}+C_{1}, \quad \text { or } \quad \frac{d y}{d x}=\sqrt{y^{4}+C_{1}} .
$$

Separating variables, we find

$$
x+C_{2}=\int\left(y^{4}+C_{1}\right)^{-1 / 2} d y
$$

In the right-hand side of the last equation, we have an integral of a differential binomial. Here $m=0, n=4, p=-\frac{1}{2}$, i.e., a non-integrable case.

Therefore, this integral cannot be expressed as a finite combination of elementary functions. However, if we use the initial conditions, we get $C_{1}=0$. So $\frac{d y}{d x}=y^{2}$, from which, considering the initial conditions, we finally find $y=\frac{1}{1-x}$.