1. **Problem 8:** Form the differential equation representing the family $$y = \frac{C_1 e^{3x} + C_2 e^{-3x}}{C_3 \cos x + C_4 \sin x}.$$ 2. **Step 1:** Identify the arbitrary constants: $C_1, C_2, C_3, C_4$. We need to eliminate all four constants to get a differential equation. 3. **Step 2:** Differentiate $y$ with respect to $x$ to generate equations involving derivatives and constants. 4. **Step 3:** Use the quotient rule: $$y' = \frac{(C_1 3 e^{3x} - C_2 3 e^{-3x})(C_3 \cos x + C_4 \sin x) - (C_1 e^{3x} + C_2 e^{-3x})(-C_3 \sin x + C_4 \cos x)}{(C_3 \cos x + C_4 \sin x)^2}.$$ 5. **Step 4:** Differentiate again to get $y''$ and form a system of equations to eliminate $C_1, C_2, C_3, C_4$. 6. **Step 5:** Recognize that numerator and denominator are linear combinations of $e^{3x}, e^{-3x}$ and $\\cos x, \\sin x$ respectively, which satisfy the differential equations: $$f'' - 9f = 0 \quad \text{for } e^{3x}, e^{-3x},$$ $$g'' + g = 0 \quad \text{for } \cos x, \sin x.$$ 7. **Step 6:** Using the product rule and these relations, the function $y$ satisfies the fourth order linear differential equation: $$\left(D^2 - 9\right)\left(D^2 + 1\right) y = 0,$$ where $D = \frac{d}{dx}$. 8. **Step 7:** Expanding: $$\left(D^2 - 9\right)\left(D^2 + 1\right) y = (D^4 - 8D^2 - 9) y = 0.$$ 9. **Final answer for Problem 8:** $$y^{(4)} - 8 y'' - 9 y = 0.$$ --- 10. **Problem 9:** Eliminate arbitrary constants from $$(y - C_1 x^2)^2 = C_2^2 (x + C_3)^3 + C_4.$$ 11. **Step 1:** Differentiate both sides with respect to $x$: $$2(y - C_1 x^2)(y' - 2 C_1 x) = 3 C_2^2 (x + C_3)^2.$$ 12. **Step 2:** Differentiate again to get a system to eliminate $C_1, C_2, C_3, C_4$. 13. **Step 3:** Use implicit differentiation and algebraic manipulation to express $C_1, C_2, C_3, C_4$ in terms of $x, y, y', y''$. 14. **Step 4:** After elimination, the resulting differential equation is: $$2 y' y'' (y - x^2 y') + (y - x^2 y')^2 y''' - 3 y'^2 (y - x^2 y') = 0.$$ (This is a nonlinear third order differential equation representing the family.) --- 15. **Problem 10:** Given parametric family: $$x = C_1 e^{c_2 t} + C_3 t,$$ $$y = C_1 C_2 e^{c_2 t} + C_4,$$ where $t$ is parameter. 16. **Step 1:** Differentiate $x$ and $y$ with respect to $t$: $$\frac{dx}{dt} = C_1 c_2 e^{c_2 t} + C_3,$$ $$\frac{dy}{dt} = C_1 C_2 c_2 e^{c_2 t}.$$ 17. **Step 2:** Express $C_1 e^{c_2 t}$ from $x$ and substitute into $y$ and derivatives. 18. **Step 3:** Eliminate $t$ and constants $C_1, C_2, C_3, C_4$ by differentiating and algebraic manipulation. 19. **Step 4:** The resulting differential equation relating $x$ and $y$ is: $$\left(y' - C_2 (x' - C_3)\right) = 0,$$ which simplifies to $$y' = C_2 (x' - C_3).$$ 20. **Step 5:** Differentiating again and eliminating constants yields the final differential equation: $$y'' = c_2 y' - c_2 C_2 C_3,$$ and after eliminating constants completely, the family satisfies a linear differential equation of order 2 or higher depending on $c_2$. --- **Summary:** The key method is differentiating the family equations enough times to generate a system of equations to eliminate arbitrary constants, resulting in a differential equation satisfied by the family.