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The exercise you've shared is focused on deriving the differential equations by eliminating the arbitrary constants present in the given solutions. Let's go through each problem step by step to determine the corresponding differential equation.

### 1. \( (y - 5)^2 = cx \)

- Differentiate both sides with respect to \( x \):
  \[
  2(y - 5) \frac{dy}{dx} = c
  \]
  Rearranging this, we get:
  \[
  \frac{dy}{dx} = \frac{c}{2(y - 5)}
  \]
  The constant \( c \) can be eliminated using the original equation \( (y - 5)^2 = cx \), giving:
  \[
  c = \frac{(y - 5)^2}{x}
  \]
  Substituting this back into the derivative:
  \[
  \frac{dy}{dx} = \frac{(y - 5)}{2x}
  \]
  Multiplying both sides by \( 2x \), we get the final differential equation:
  \[
  2x \frac{dy}{dx} = y - 5
  \]

### 2. \( y = c_1 + c_2 x + c_3 x^2 + c_4 x^3 \)

- Since there are 4 arbitrary constants, we'll need to differentiate multiple times to eliminate them.
  1. First derivative:
     \[
     \frac{dy}{dx} = c_2 + 2c_3 x + 3c_4 x^2
     \]
  2. Second derivative:
     \[
     \frac{d^2 y}{dx^2} = 2c_3 + 6c_4 x
     \]
  3. Third derivative:
     \[
     \frac{d^3 y}{dx^3} = 6c_4
     \]
  4. Fourth derivative:
     \[
     \frac{d^4 y}{dx^4} = 0
     \]
  So, the required differential equation is:
  \[
  \frac{d^4 y}{dx^4} = 0
  \]

### 3. \( c x^2 + y^2 = 4 \)

- Differentiate both sides with respect to \( x \):
  \[
  2cx + 2y \frac{dy}{dx} = 0
  \]
  Simplify this to:
  \[
  cx + y \frac{dy}{dx} = 0
  \]
  This is the required differential equation.

### 4. \( y = c_1 e^{-x} + c_2 e^{2x} + c_3 e^x \)

- Differentiate multiple times to eliminate constants:
  1. First derivative:
     \[
     \frac{dy}{dx} = -c_1 e^{-x} + 2c_2 e^{2x} + c_3 e^x
     \]
  2. Second derivative:
     \[
     \frac{d^2 y}{dx^2} = c_1 e^{-x} + 4c_2 e^{2x} + c_3 e^x
     \]
  3. Third derivative:
     \[
     \frac{d^3 y}{dx^3} = -c_1 e^{-x} + 8c_2 e^{2x} + c_3 e^x
     \]
  4. Fourth derivative:
     \[
     \frac{d^4 y}{dx^4} = c_1 e^{-x} + 16c_2 e^{2x} + c_3 e^x
     \]
  So, the differential equation is:
  \[
  \frac{d^4 y}{dx^4} - 3 \frac{d^3 y}{dx^3} - 3 \frac{d^2 y}{dx^2} + \frac{dy}{dx} = 0
  \]

Would you like more details on any of these steps, or help with further clarification?

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### Related Questions:
1. How can we apply the method of arbitrary constant elimination to a system of differential equations?
2. What is the significance of higher-order derivatives in eliminating constants?
3. How do the constants in solutions to differential equations represent initial conditions?
4. What are the methods to solve non-homogeneous differential equations?
5. How do boundary conditions affect the solution of a differential equation?

**Tip:** When eliminating arbitrary constants, always consider how many constants you have and differentiate enough times to remove all of them.

Determine the differential equation whose solution is given by: 1. (y - 5)^2 = cx, 2. y = c1 + c2x + c3x^2 + c4x^3, 3. cx^2 + y^2 = 4, 4. y = c1e^(-x) + c2e^(2x) + c3e^(x).

Learn how to derive differential equations by eliminating arbitrary constants from given solutions. This includes problems like (y - 5)^2 = cx and y = c1 + c2x + c3x^2 + c4x^3. Suitable for students studying undergraduate calculus or differential equations.

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