Understanding Differential Equation Solution Intervals

The question asks us to identify the correct interval where a unique solution exists for a given first-order linear differential equation with a specified initial condition. The existence and uniqueness of solutions for such equations are determined by the continuity of the coefficients involved, as stated by the Existence and Uniqueness Theorem.

Applying the Existence and Uniqueness Theorem

The theorem applies to linear differential equations of the form:

$$ \frac{{dy}}{{dt}} + P(t)y = Q(t) $$

The theorem guarantees that if \(P(t)\) and \(Q(t)\) are continuous on an open interval \(I\) that contains the initial value \(t_0\), then a unique solution to the initial value problem exists on the interval \(I\).

Step-by-Step Analysis

1. Standardize the Equation:

   The given differential equation is:

   $$ \left( {{t^2} - 81} \right)\frac{{dy}}{{dt}} + 5ty = \sin \left( t \right) $$

   To find \(P(t)\) and \(Q(t)\), we first divide by the coefficient of \(\frac{dy}{dt}\), which is \(\left( {{t^2} - 81} \right)\). We must assume \(t^2 - 81 \neq 0\):

   $$ \frac{{dy}}{{dt}} + \frac{{5t}}{{{t^2} - 81}}y = \frac{{\sin \left( t \right)}}{{{t^2} - 81}} $$

2. Identify \(P(t)\) and \(Q(t)\):

   By comparing the standardized equation with the standard form \(\frac{{dy}}{{dt}} + P(t)y = Q(t)\), we identify:

   • \( P(t) = \frac{{5t}}{{{t^2} - 81}} \)
   • \( Q(t) = \frac{{\sin \left( t \right)}}{{{t^2} - 81}} \)
3. Determine Points of Discontinuity:

   The continuity of \(P(t)\) and \(Q(t)\) depends on the denominator \(t^2 - 81\). This expression is zero when:

   $$ t^2 - 81 = 0 $$

   $$ t^2 = 81 $$

   $$ t = \pm \sqrt{81} $$

   $$ t = 9 \quad \text{and} \quad t = -9 $$

   These are the points where the coefficients might be discontinuous. They divide the real number line into three intervals: \((-\infty, -9)\), \((-9, 9)\), and \((9, \infty)\). Within each of these intervals, \(P(t)\) and \(Q(t)\) are continuous.

4. Locate the Initial Condition:

   The initial condition is given as \(y(1) = 2\pi\). The value \(t_0 = 1\) is the point where the solution must exist.

5. Find the Relevant Interval:

   We need to find the interval from the points of discontinuity (\(-9, 9\)) that contains the initial condition \(t_0 = 1\). The value \(t=1\) clearly falls within the interval \((-9, 9)\).

   Therefore, according to the Existence and Uniqueness Theorem, a unique solution is guaranteed to exist on the interval \((-9, 9)\).

6. Evaluate the Options:

   Now, we check the given options against our findings:

   • Option 1: \((–2, 2)\): This interval contains \(t_0 = 1\). Since \((–2, 2)\) is a sub-interval of \((-9, 9)\), both \(P(t)\) and \(Q(t)\) are continuous throughout \((–2, 2)\). Thus, a unique solution exists here.
   • Option 2: \((–10, 10)\): This interval contains \(t_0 = 1\), but it extends beyond the points of discontinuity \(t = -9\) and \(t = 9\). The coefficients are not continuous over the entire interval \((–10, 10)\), so the theorem doesn't guarantee a unique solution on this whole interval.
   • Option 3: \((–10, 2)\): This interval contains \(t_0 = 1\), but it includes the point of discontinuity \(t = -9\). The coefficients are not continuous over the entire interval \((–10, 2)\).
   • Option 4: \((0, 10)\): This interval contains \(t_0 = 1\), but it includes the point of discontinuity \(t = 9\). The coefficients are not continuous over the entire interval \((0, 10)\).

   Only Option 1, \((–2, 2)\), represents an interval containing the initial condition \(t_0=1\) where the coefficients \(P(t)\) and \(Q(t)\) are guaranteed to be continuous.

Final Conclusion

The interval \((–2, 2)\) satisfies the conditions of the Existence and Uniqueness Theorem for the given differential equation and initial condition, as it contains \(t_0 = 1\) and lies entirely within the region \((-9, 9)\) where the coefficients \(P(t)\) and \(Q(t)\) are continuous.