The problem is to solve the differential equation $y''' = 20x^3$.

1. Problem Description

The problem is to solve the differential equation $y''' = 20x^3$.

2. Solution Steps

We need to integrate the equation $y''' = 20x^3$ three times to find $y$.

First integration:

$\int y''' dx = \int 20x^3 dx$

$y'' = 20 \cdot \frac{x^4}{4} + C_1$

$y'' = 5x^4 + C_1$

Second integration:

$\int y'' dx = \int (5x^4 + C_1) dx$

$y' = 5 \cdot \frac{x^5}{5} + C_1x + C_2$

$y' = x^5 + C_1x + C_2$

Third integration:

$\int y' dx = \int (x^5 + C_1x + C_2) dx$

$y = \frac{x^6}{6} + C_1 \cdot \frac{x^2}{2} + C_2x + C_3$

$y = \frac{1}{6}x^6 + \frac{C_1}{2}x^2 + C_2x + C_3$

We can replace $\frac{C_1}{2}$ with a new constant $A$, then $C_2$ with $B$, and $C_3$ with $C$.

$y = \frac{1}{6}x^6 + Ax^2 + Bx + C$

3. Final Answer

$y = \frac{1}{6}x^6 + Ax^2 + Bx + C$