The given first-order linear differential equation is:

$y' + \frac{t}{t^2 - 4}y = \frac{e^t}{t - 6}$

with the initial condition $y(0) = 0$.

Step 1: Rewrite the equation in standard form

We can rewrite the equation in the standard form:

$y' + p(t)y = g(t)$

where:

• $p(t) = \frac{t}{t^2 - 4}$
• $g(t) = \frac{e^t}{t - 6}$

Step 2: Determine singularities of $p(t)$ and $g(t)$

The existence and uniqueness theorem for first-order linear differential equations guarantees the existence and uniqueness of a solution on an interval where $p(t)$ and $g(t)$ are continuous. Therefore, we need to check where $p(t)$ and $g(t)$ are discontinuous.

Analyze $p(t) = \frac{t}{t^2 - 4}$

The function $p(t)$ has singularities where the denominator $t^2 - 4 = 0$, i.e., at:

$t^2 = 4 \quad \Rightarrow \quad t = \pm 2$

Thus, $p(t)$ is discontinuous at $t = 2$ and $t = -2$.

Analyze $g(t) = \frac{e^t}{t - 6}$

The function $g(t)$ has a singularity at $t = 6$ because the denominator $t - 6 = 0$ when $t = 6$.

Step 3: Determine the largest interval containing $t = 0$

To guarantee existence and uniqueness of the solution, the initial condition $y(0) = 0$ must be within an interval where both $p(t)$ and $g(t)$ are continuous. The singularities are at $t = -2$, $t = 2$, and $t = 6$. The largest interval around $t = 0$ that does not include any of these singularities is:

$(-2, 2)$

Thus, the largest interval on which the existence and uniqueness theorem guarantees a unique solution is $(-2, 2)$.

Conclusion:

The largest interval on which the existence and uniqueness theorem guarantees the existence of a unique solution is $(-2, 2)$.

Would you like further details or clarifications? Here are some related questions:

1. What happens if the initial condition is chosen at $t = 2$?
2. How does the form of the initial value problem change near singularities?
3. How do singularities affect the behavior of the solution?
4. How would you solve this differential equation using an integrating factor?
5. What methods can be used to extend the solution beyond the interval $(-2, 2)$?

Tip: When dealing with first-order differential equations, always identify discontinuities in the coefficients to apply the existence and uniqueness theorem correctly.