The image contains four exercises asking to determine the differential equations from the given solutions. These solutions contain arbitrary constants that must be eliminated to find the associated differential equations. Here are the given problems:

1. $(y - 5)^2 = cx$
2. $y = c_1 + c_2x + c_3x^2 + c_4x^3$
3. $cx^2 + y^2 = 4$
4. $y = c_1e^x + c_2e^{-x} + c_3e^{3x}$

Solving these exercises:

I'll guide through the elimination of arbitrary constants for each solution.

1. $(y - 5)^2 = cx$

• Differentiate both sides with respect to $x$ to eliminate $c$: $2(y - 5) \frac{dy}{dx} = c$ Differentiate again to remove $c$: $2\left( \frac{dy}{dx} \right)^2 + 2(y - 5) \frac{d^2y}{dx^2} = 0$ So the differential equation becomes: $(y - 5) \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0$

2. $y = c_1 + c_2x + c_3x^2 + c_4x^3$

• Differentiate three times to remove the constants $c_1$, $c_2$, $c_3$, and $c_4$:
  1. First derivative: $\frac{dy}{dx} = c_2 + 2c_3x + 3c_4x^2$
  2. Second derivative: $\frac{d^2y}{dx^2} = 2c_3 + 6c_4x$
  3. Third derivative: $\frac{d^3y}{dx^3} = 6c_4$ Since the third derivative is a constant, this implies that the differential equation is: $\frac{d^3y}{dx^3} = 0$

3. $cx^2 + y^2 = 4$

• Differentiate both sides with respect to $x$: $2cx + 2y \frac{dy}{dx} = 0$ Simplifying, we get: $cx + y \frac{dy}{dx} = 0$ Differentiate again: $c + \left( \frac{dy}{dx} \right)^2 + y \frac{d^2y}{dx^2} = 0$ So, the differential equation becomes: $c + \left( \frac{dy}{dx} \right)^2 + y \frac{d^2y}{dx^2} = 0$

4. $y = c_1e^x + c_2e^{-x} + c_3e^{3x}$

• Differentiate three times:
  1. First derivative: $\frac{dy}{dx} = c_1e^x - c_2e^{-x} + 3c_3e^{3x}$
  2. Second derivative: $\frac{d^2y}{dx^2} = c_1e^x + c_2e^{-x} + 9c_3e^{3x}$
  3. Third derivative: $\frac{d^3y}{dx^3} = c_1e^x - c_2e^{-x} + 27c_3e^{3x}$ So, the general third derivative is: $\frac{d^3y}{dx^3} - 3\frac{dy}{dx} = 0$ Therefore, the differential equation is: $\frac{d^3y}{dx^3} - 3\frac{dy}{dx} = 0$

Would you like any further clarification on these steps? Let me know if you need more details or have any questions.

Here are 5 questions to expand your understanding:

1. How does differentiating help eliminate arbitrary constants in a solution?
2. Can you identify other methods to eliminate constants besides differentiation?
3. Why does the third derivative in question 2 lead to a simple result?
4. What happens if the arbitrary constants aren't fully eliminated in a differential equation?
5. How would the solutions differ if initial conditions were provided?

Tip: Always check if the resulting equation is homogeneous when differentiating multiple times to eliminate constants.