Question

Solve the following differential equation:
y'' - x = 0

Answer 1

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.