To solve the differential equation \( y'' - x = 0 \), we start by rearranging it into a more standard form:
\[ y'' = x \]
Next, we integrate both sides with respect to \( x \):
\[ \int y'' , dx = \int x , dx \]
The left side integrates to the first derivative of \( y \):
\[ y' = \frac{x^2}{2} + C_1 \]
where \( C_1 \) is the constant of integration. Now, we integrate again to find \( y \):
\[ \int y' , dx = \int \left( \frac{x^2}{2} + C_1 \right) dx \]
This results in:
\[ y = \frac{x^3}{6} + C_1 x + C_2 \]
where \( C_2 \) is another constant of integration. Therefore, the general solution to the differential equation \( y'' - x = 0 \) is:
\[ y = \frac{x^3}{6} + C_1 x + C_2 \]
Here, \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions or boundary conditions if they are provided.
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