We are given the system of differential equations:

$x' + 6y = 0 \quad \text{(1)}$ $x' - y' = 0 \quad \text{(2)}$

Our goal is to use the elimination method to find a general solution.

Step 1: Solve for $x'$ from one of the equations

From equation (2), we have: $x' = y' \quad \text{(3)}$

Step 2: Substitute $x'$ from equation (3) into equation (1)

Substitute $x' = y'$ into equation (1): $y' + 6y = 0$

This is a first-order linear differential equation for $y$.

Step 3: Solve the differential equation for $y$

The equation $y' + 6y = 0$ is separable, and can be rewritten as: $\frac{dy}{dt} = -6y$

Separating variables and integrating: $\int \frac{1}{y} \, dy = \int -6 \, dt$

This gives: $\ln |y| = -6t + C_1$

Exponentiating both sides: $|y| = e^{-6t + C_1} = e^{C_1} e^{-6t}$

Let $C_2 = e^{C_1}$, then: $y(t) = C_2 e^{-6t}$

Step 4: Solve for $x$ using equation (3)

From equation (3), $x' = y'$. We know that: $y' = -6C_2 e^{-6t}$

Therefore, $x' = -6C_2 e^{-6t}$. Now integrate this to find $x$: $x(t) = \int -6C_2 e^{-6t} \, dt = C_2 e^{-6t} + C_3$

General Solution

Thus, the general solution for the system is: $x(t) = C_2 e^{-6t} + C_3$ $y(t) = C_2 e^{-6t}$

Do you want further details or have any questions about these steps?

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Here are 5 related questions:

1. How can we verify that this solution satisfies both differential equations?
2. What is the significance of the constants $C_2$ and $C_3$ in the solution?
3. How would the system change if the coefficient of $y$ in the first equation was different?
4. Can we solve this system using matrix methods instead of elimination?
5. What happens to the solution as $t \to \infty$?

Tip: When solving linear differential equations, always check if you can simplify the system using substitution or elimination methods to reduce the complexity.