1. **State the problem:** We are given the differential equation $$y'' = 32^3$$ with initial conditions $$y(4) = 1$$ and $$y'(4) = 4$$. We need to find the function $$y(x)$$. 2. **Recall the formula and rules:** The equation $$y'' = C$$ where $$C$$ is a constant means the second derivative of $$y$$ is constant. To find $$y$$, integrate twice. 3. **First integration:** Integrate $$y'' = 32^3$$ with respect to $$x$$ to find $$y'$$: $$y' = \int 32^3 \, dx = 32^3 x + C_1$$ where $$C_1$$ is a constant of integration. 4. **Apply initial condition for $$y'$$:** Given $$y'(4) = 4$$, $$4 = 32^3 \cdot 4 + C_1 \implies C_1 = 4 - 4 \cdot 32^3$$ 5. **Second integration:** Integrate $$y'$$ to find $$y$$: $$y = \int (32^3 x + C_1) \, dx = \frac{32^3}{2} x^2 + C_1 x + C_2$$ where $$C_2$$ is another constant of integration. 6. **Apply initial condition for $$y$$:** Given $$y(4) = 1$$, $$1 = \frac{32^3}{2} \cdot 4^2 + C_1 \cdot 4 + C_2$$ Substitute $$C_1$$ from step 4: $$1 = \frac{32^3}{2} \cdot 16 + (4 - 4 \cdot 32^3) \cdot 4 + C_2$$ Simplify: $$1 = 8 \cdot 32^3 + 16 - 16 \cdot 32^3 + C_2$$ $$1 = 16 - 8 \cdot 32^3 + C_2$$ $$C_2 = 1 - 16 + 8 \cdot 32^3 = -15 + 8 \cdot 32^3$$ 7. **Final solution:** $$y = \frac{32^3}{2} x^2 + \left(4 - 4 \cdot 32^3\right) x + \left(-15 + 8 \cdot 32^3\right)$$ This is the function $$y(x)$$ satisfying the given differential equation and initial conditions.