4.94. Solve the Cauchy problem: $y^{\prime \prime}-8 y^{3}=0 ; y(0)=-1 ; y^{\prime}(0)=2$.
Solution. Multiply both sides of the equation by $y^{\prime}$ and notice that the left side becomes a complete derivative:

$$
y^{\prime} \cdot y^{\prime \prime}-8 y^{3} \cdot y^{\prime}=\frac{d}{d x}\left(\frac{y^{\prime 2}}{2}-2 y^{4}\right)=0
$$

From this, considering the initial conditions, we find

$$
y^{\prime 2}=4 y^{4}+C_{1} \Rightarrow y^{\prime}=2 y^{2} .
$$

Integrate the last equation and consider the initial condition $y(0)=-1$ :

$$
-\frac{1}{y}=2 x+C_{2} \Rightarrow y=-\frac{1}{1+2 x}
$$

### 4.3.1.2.  PROBLEMS FOR INDEPENDENT SOLUTION WITH ANSWERS

## Integrate the equations.