1. **State the problem:** We are given the second derivative of a curve as $$\frac{d^2y}{dx^2} = 45 \cos 3x + 2 \sin x$$ with initial conditions $$\frac{dy}{dx} = -2$$ at $$x=0$$ and the curve passes through the point $$(\pi, -1)$$. We need to find the equation of the curve $$y(x)$$. 2. **Recall the formula:** To find $$y(x)$$, we integrate the second derivative twice. First, integrate $$\frac{d^2y}{dx^2}$$ to get $$\frac{dy}{dx}$$, then integrate $$\frac{dy}{dx}$$ to get $$y$$. 3. **Integrate the second derivative:** $$\frac{dy}{dx} = \int (45 \cos 3x + 2 \sin x) \, dx = 45 \int \cos 3x \, dx + 2 \int \sin x \, dx$$ 4. **Calculate each integral:** $$\int \cos 3x \, dx = \frac{\sin 3x}{3}$$ $$\int \sin x \, dx = -\cos x$$ 5. **Substitute back:** $$\frac{dy}{dx} = 45 \cdot \frac{\sin 3x}{3} + 2 (-\cos x) + C_1 = 15 \sin 3x - 2 \cos x + C_1$$ 6. **Use initial condition $$\frac{dy}{dx} = -2$$ at $$x=0$$:** $$-2 = 15 \sin 0 - 2 \cos 0 + C_1 = 0 - 2 (1) + C_1 = -2 + C_1$$ 7. **Solve for $$C_1$$:** $$-2 = -2 + C_1 \implies C_1 = 0$$ 8. **So, $$\frac{dy}{dx} = 15 \sin 3x - 2 \cos x$$** 9. **Integrate $$\frac{dy}{dx}$$ to find $$y$$:** $$y = \int (15 \sin 3x - 2 \cos x) \, dx = 15 \int \sin 3x \, dx - 2 \int \cos x \, dx + C_2$$ 10. **Calculate each integral:** $$\int \sin 3x \, dx = -\frac{\cos 3x}{3}$$ $$\int \cos x \, dx = \sin x$$ 11. **Substitute back:** $$y = 15 \left(-\frac{\cos 3x}{3}\right) - 2 \sin x + C_2 = -5 \cos 3x - 2 \sin x + C_2$$ 12. **Use the point $$(\pi, -1)$$ to find $$C_2$$:** $$-1 = -5 \cos 3\pi - 2 \sin \pi + C_2$$ 13. **Evaluate trigonometric functions:** $$\cos 3\pi = \cos (\pi) = -1$$ $$\sin \pi = 0$$ 14. **Substitute values:** $$-1 = -5 (-1) - 2 (0) + C_2 = 5 + C_2$$ 15. **Solve for $$C_2$$:** $$C_2 = -1 - 5 = -6$$ 16. **Final equation of the curve:** $$y = -5 \cos 3x - 2 \sin x - 6$$