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To find the largest interval where a unique solution is guaranteed for each given linear second-order differential equation, we apply the Existence and Uniqueness Theorem (Theorem 3.2 mentioned in the problem). The theorem generally states that for a second-order linear differential equation of the form:

\[
y'' + p(t) y' + q(t) y = g(t)
\]

with initial conditions \(y(t_0) = y_0\) and \(y'(t_0) = y_0'\), a unique solution exists as long as the functions \(p(t)\), \(q(t)\), and \(g(t)\) are continuous on an interval containing \(t_0\).

### (a) Equation:
\[
\frac{d^2 y}{dt^2} + \frac{t-1}{t^2 - 4} \frac{dy}{dt} - \frac{3}{t} y = 0
\]
with \(y(1) = -1\), \(y'(1) = 5\).

- Singularities occur where the coefficients of \(y'\) and \(y\) are undefined. 
- The term \(\frac{t-1}{t^2 - 4}\) has singularities where \(t^2 - 4 = 0\), i.e., at \(t = \pm 2\).
- Also, \(\frac{3}{t}\) has a singularity at \(t = 0\).
- Thus, the singularities are at \(t = -2, 0, 2\).
- Since the initial condition is given at \(t = 1\), the largest interval containing \(t = 1\) and excluding singularities is \((-2, 2)\).

### (b) Equation:
\[
(t + 2)\frac{d^2 y}{dt^2} + 2 \ln |t| \frac{dy}{dt} - 3y = 0
\]
with \(y(-1) = 0\), \(y'(-1) = 1\).

- The coefficient \((t+2)\) introduces a singularity at \(t = -2\), and \(\ln |t|\) introduces a singularity at \(t = 0\) because \(\ln |t|\) is undefined at \(t = 0\).
- Thus, the singularities are at \(t = -2\) and \(t = 0\).
- Since the initial condition is given at \(t = -1\), the largest interval containing \(t = -1\) and excluding singularities is \((-2, 0)\).

### (c) Equation:
\[
t \frac{d^2 y}{dt^2} + (t-1)\frac{dy}{dt} - \tan(t) y = 0
\]
with \(y(1) = 1\), \(y'(1) = -1\).

- The coefficient \(\tan(t)\) introduces singularities where \(\tan(t)\) is undefined, i.e., at \(t = \frac{\pi}{2} + n\pi\) for \(n \in \mathbb{Z}\).
- The term \(t \frac{d^2 y}{dt^2}\) introduces a singularity at \(t = 0\).
- Thus, singularities are at \(t = 0\), \(\frac{\pi}{2}\), \(\frac{3\pi}{2},\) etc.
- Since the initial condition is given at \(t = 1\), the largest interval containing \(t = 1\) and excluding singularities is \((0, \frac{\pi}{2})\).

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Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What conditions guarantee the continuity of coefficients in linear differential equations?
2. How does the Existence and Uniqueness Theorem apply to nonlinear differential equations?
3. What happens if an initial value is set at a singular point of the differential equation?
4. How can the interval of existence change if higher-order differential equations are considered?
5. Why do transcendental functions like \(\tan(t)\) introduce multiple singularities in differential equations?

**Tip**: Always check the points of discontinuity or undefined behavior in the coefficient functions when applying the Existence and Uniqueness Theorem.

Find the largest interval on which a unique solution is guaranteed to each higher order initial value problem by Theorem 3.2 for linear DEs.

Explore how to determine the largest interval where a unique solution is guaranteed for linear differential equations using Theorem 3.2. This process involves identifying singularities and checking the continuity of functions in the equation.

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