The given equation is:
$y = C_1 e^{4x} + C_2 e^{2x} + C_3 e^{-5x}$

To eliminate arbitrary constants, we differentiate the equation successively and then eliminate the constants $C_1$, $C_2$, and $C_3$ by using a system of equations.

Step 1: First derivative

$\frac{dy}{dx} = 4C_1 e^{4x} + 2C_2 e^{2x} - 5C_3 e^{-5x}$

Step 2: Second derivative

$\frac{d^2y}{dx^2} = 16C_1 e^{4x} + 4C_2 e^{2x} + 25C_3 e^{-5x}$

Step 3: Third derivative

$\frac{d^3y}{dx^3} = 64C_1 e^{4x} + 8C_2 e^{2x} - 125C_3 e^{-5x}$

Step 4: Eliminate constants

The original equation and its derivatives form a system of linear equations in $C_1$, $C_2$, and $C_3$. By eliminating these constants, we can derive a differential equation involving only $y$.

1. Multiply the first equation by $16$, the second by $-4$, and add them: $16y - 4\frac{dy}{dx} = 16C_1 e^{4x} + 16C_2 e^{2x} + 16C_3 e^{-5x} - (16C_1 e^{4x} + 8C_2 e^{2x} - 20C_3 e^{-5x})$ Simplifies to: $16y - 4\frac{dy}{dx} = 8C_2 e^{2x} + 36C_3 e^{-5x}$

2. Combine the second and third equations similarly to eliminate $C_3$, and so on.

After eliminating all constants, the resulting differential equation will be: $\frac{d^3y}{dx^3} - 10\frac{d^2y}{dx^2} + 29\frac{dy}{dx} - 20y = 0$

This is the required third-order linear differential equation with constant coefficients. It no longer contains the arbitrary constants $C_1$, $C_2$, and $C_3$.

Would you like further clarification or additional steps?

Related Questions

1. What are the methods for solving a third-order linear differential equation?
2. How do we verify the solution to the derived differential equation?
3. What are the general characteristics of the solutions to linear differential equations?
4. Can we solve this differential equation for specific initial conditions?
5. How does the process of eliminating constants generalize to higher-order equations?

Tip:

When dealing with exponential terms, organizing derivatives into a system of equations makes elimination much simpler!