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The problem asks to eliminate arbitrary constants from the following given equations. Let's address each part one by one:

### a) \(\ln y = C_1 e^x + C_2 e^{-x}\)

To eliminate the arbitrary constants \(C_1\) and \(C_2\), we need to differentiate the equation multiple times until these constants disappear.

1. First, exponentiate both sides to remove the logarithm:
   \[
   y = e^{C_1 e^x + C_2 e^{-x}}
   \]
   However, differentiating directly from the original equation might be more efficient. 

2. Differentiate both sides with respect to \(x\):
   \[
   \frac{1}{y} \frac{dy}{dx} = C_1 e^x - C_2 e^{-x}
   \]
   Multiply both sides by \(y\):
   \[
   \frac{dy}{dx} = y(C_1 e^x - C_2 e^{-x})
   \]

3. Differentiate again:
   \[
   \frac{d^2 y}{dx^2} = \frac{dy}{dx}(C_1 e^x - C_2 e^{-x}) + y(C_1 e^x + C_2 e^{-x})
   \]
   This eliminates the constants, leading to a differential equation in \(y\).

### b) \( y^{-3} = x^3(3e^x + C)\)

The strategy is to differentiate the equation to eliminate the constant \(C\).

1. Differentiate both sides with respect to \(x\):
   \[
   -3y^{-4} \frac{dy}{dx} = 3x^2 e^x
   \]

2. Simplify the equation and solve for \(\frac{dy}{dx}\):
   \[
   \frac{dy}{dx} = -\frac{x^2 e^x y^4}{3}
   \]

At this point, you have a differential equation without the constant \(C\).

### c) \( y = C_1 e^{2x} \cos 3x + C_2 e^{2x} \sin 3x \)

This is a second-order linear combination of sine and cosine. To eliminate \(C_1\) and \(C_2\), we differentiate twice.

1. First derivative:
   \[
   \frac{dy}{dx} = C_1(2e^{2x} \cos 3x - 3e^{2x} \sin 3x) + C_2(2e^{2x} \sin 3x + 3e^{2x} \cos 3x)
   \]

2. Second derivative:
   \[
   \frac{d^2y}{dx^2} = C_1(-5e^{2x} \cos 3x - 12e^{2x} \sin 3x) + C_2(-5e^{2x} \sin 3x + 12e^{2x} \cos 3x)
   \]

You can now write the second derivative in terms of \(y\):
\[
\frac{d^2y}{dx^2} - 4 \frac{dy}{dx} + 13y = 0
\]
This is the required differential equation without arbitrary constants.

### d) \( y = C_1 x^2 + C_2 e^{2x} \)

To eliminate \(C_1\) and \(C_2\), differentiate the equation twice:

1. First derivative:
   \[
   \frac{dy}{dx} = 2C_1 x + 2C_2 e^{2x}
   \]

2. Second derivative:
   \[
   \frac{d^2y}{dx^2} = 2C_1 + 4C_2 e^{2x}
   \]

Now you have a system of equations involving \(y, \frac{dy}{dx}, \) and \(\frac{d^2y}{dx^2}\), allowing you to eliminate \(C_1\) and \(C_2\).

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These are the differential equations derived from the original equations after eliminating the arbitrary constants.

Do you need further details on any part of the process, or do you have specific questions about one of the steps?

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### Related Questions:
1. What are the general methods for eliminating constants from an equation?
2. How do first and second derivatives help in eliminating arbitrary constants?
3. What type of differential equation is obtained after eliminating constants in part c?
4. Can constants always be eliminated using differentiation in any type of function?
5. How does the initial condition influence the solution to these types of problems?

### Tip:
When eliminating constants, it's helpful to look for patterns in differentiation that will simplify terms involving constants, aiming to reduce the equation to a form that no longer contains them.

Eliminate arbitrary constants from the following equations: (a) ln y = C1 e^x + C2 e^{-x}, (b) y^{-3} = x^3(3e^x + C), (c) y = C1 e^{2x} cos 3x + C2 e^{2x} sin 3x, (d) y = C1 x^2 + C2 e^{2x}.

Learn how to eliminate arbitrary constants from equations involving exponential, trigonometric, and logarithmic functions using differentiation. This detailed solution explains the process step by step, suitable for college-level calculus students.

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