Solve the initial value problem. \[ \frac{d^{2} y}{d x^{2}}=12-30 x, y^{\prime}(0)=7 \text {, and } y(0)=7 \] \( y(x)= \)

To solve the initial value problem given by the second-order differential equation \[ \frac{d^{2} y}{d x^{2}} = 12 - 30x, \] we will first integrate the equation to find \( y'(x) \). ### Step 1: Integrate the second derivative Integrating the right-hand side with respect to \( x \): \[ \frac{dy}{dx} = \int (12 - 30x) \, dx = 12x - 15x^2 + C_1, \] where \( C_1 \) is a constant of integration. ### Step 2: Integrate the first derivative Next, we integrate \( y'(x) \) to find \( y(x) \): \[ y(x) = \int (12x - 15x^2 + C_1) \, dx = 6x^2 - 5x^3 + C_1 x + C_2, \] where \( C_2 \) is another constant of integration. ### Step 3: Apply initial conditions We have the initial conditions \( y(0) = 7 \) and \( y'(0) = 7 \). 1. **Using \( y(0) = 7 \)**: Substituting \( x = 0 \) into \( y(x) \): \[ y(0) = 6(0)^2 - 5(0)^3 + C_1(0) + C_2 = C_2 = 7. \] So, we have \( C_2 = 7 \). 2. **Using \( y'(0) = 7 \)**: Substituting \( x = 0 \) into \( y'(x) \): \[ y'(0) = 12(0) - 15(0)^2 + C_1 = C_1 = 7. \] So, we have \( C_1 = 7 \). ### Step 4: Write the final solution Now we can substitute \( C_1 \) and \( C_2 \) back into the expression for \( y(x) \): \[ y(x) = 6x^2 - 5x^3 + 7x + 7. \] Thus, the solution to the initial value problem is \[ \boxed{y(x) = -5x^3 + 6x^2 + 7x + 7}. \]