To analyze the system of differential equations given by:

1. $\frac{dx}{dt} = -2x$
2. $\frac{dy}{dt} = -8x - 6y$

Step 1: Solve for $x(t)$

The first equation can be solved directly:

$\frac{dx}{dt} = -2x$

This is a separable equation. Rearranging gives:

$\frac{1}{x}dx = -2dt$

Integrating both sides:

$\ln |x| = -2t + C \implies x(t) = C_1 e^{-2t}$

where $C_1 = e^C$ is a constant.

Step 2: Substitute $x(t)$ into the equation for $y(t)$

Now, substitute $x(t)$ into the second equation:

$\frac{dy}{dt} = -8x - 6y$

Substituting $x(t)$:

$\frac{dy}{dt} = -8(C_1 e^{-2t}) - 6y$

This is a linear first-order differential equation in $y$. To solve it, we can use an integrating factor. The equation can be rewritten as:

$\frac{dy}{dt} + 6y = -8C_1 e^{-2t}$

The integrating factor is:

$\mu(t) = e^{\int 6 dt} = e^{6t}$

Multiplying through by the integrating factor:

$e^{6t} \frac{dy}{dt} + 6e^{6t} y = -8C_1 e^{4t}$

The left-hand side is the derivative of $e^{6t} y$:

$\frac{d}{dt}(e^{6t} y) = -8C_1 e^{4t}$

Integrating both sides gives:

$e^{6t} y = -2C_1 e^{4t} + C_2$

Thus,

$y(t) = -2C_1 e^{-2t} + C_2 e^{-6t}$

Step 3: Analyze the solutions

1. For $x(t)$: As $t \to \infty$, $x(t) = C_1 e^{-2t} \to 0$.
2. For $y(t)$: As $t \to \infty$,
   • The term $-2C_1 e^{-2t} \to 0$.
   • The term $C_2 e^{-6t} \to 0$.

Thus, both $x(t)$ and $y(t)$ tend monotonically to zero.

Conclusion

The correct answer is:

(a) x and y both tend monotonically to zero.

Questions for Further Exploration:

1. What are the implications of the decay rates in each equation?
2. How would the system behave if the coefficients were positive?
3. Can you derive the solutions using a different method, such as eigenvalues?
4. What is the stability of the fixed points in this system?
5. How would the system change if we introduced a nonlinear term?

Tip

When solving systems of differential equations, always look for separation of variables or integrating factors to simplify the process.