Find the function \( F \) that satisfies the following differential equation and initial conditions \( F^{\prime \prime}(x)=1, F^{\prime}(0)=13, F(0)=4 \) The function is \( F(x)=\square \)

To solve the differential equation \( F^{\prime \prime}(x) = 1 \), we first integrate it once to find \( F^{\prime}(x) \). Integrating \( F^{\prime \prime}(x) = 1 \): \[ F^{\prime}(x) = x + C_1 \] where \( C_1 \) is a constant of integration. Next, we can integrate \( F^{\prime}(x) \): \[ F(x) = \frac{x^2}{2} + C_1 x + C_2 \] where \( C_2 \) is another constant of integration. We need to apply the initial conditions to determine \( C_1 \) and \( C_2 \). Using the first initial condition \( F^{\prime}(0) = 13 \): \[ F^{\prime}(0) = 0 + C_1 = 13 \implies C_1 = 13. \] Now we substitute \( C_1 \) back into our equation for \( F(x) \): \[ F(x) = \frac{x^2}{2} + 13x + C_2. \] Now, we use the second initial condition \( F(0) = 4 \): \[ F(0) = 0 + 0 + C_2 = 4 \implies C_2 = 4. \] Putting everything together, we find: \[ F(x) = \frac{x^2}{2} + 13x + 4. \] So, the function is \( F(x) = \frac{x^2}{2} + 13x + 4 \).