Let a be some positive parameter. Solve the initial value problem, x(1) = 1, for

x′(t) = (ax/t)+ 1.

What is the maximum interval of existence of this solution? (Be careful: there exist(s) some special
value(s) of the parameter for which the solution has a different form than for the other values.) find the general solution. 

We are tasked with solving the initial value problem (IVP):

\[
x'(t) = \frac{a x(t)}{t} + 1, \quad x(1) = 1
\]

where \( a \) is a positive parameter. Let's solve this step by step.

### Step 1: Homogeneous part of the equation

First, consider the homogeneous part of the differential equation:

\[
x'(t) = \frac{a x(t)}{t}
\]

This is a first-order linear differential equation. To solve it, we can separate variables:

\[
\frac{x'(t)}{x(t)} = \frac{a}{t}
\]

Integrating both sides:

\[
\int \frac{1}{x(t)} \, dx = \int \frac{a}{t} \, dt
\]

\[
\ln |x(t)| = a \ln |t| + C
\]

Exponentiating both sides:

\[
x(t) = C_1 t^a
\]

where \( C_1 = e^C \) is the constant of integration.

### Step 2: General solution of the non-homogeneous equation

Now, let's solve the non-homogeneous equation:

\[
x'(t) = \frac{a x(t)}{t} + 1
\]

We will use the **method of integrating factors**. Rewrite the equation in standard form:

\[
x'(t) - \frac{a}{t} x(t) = 1
\]

The integrating factor is:

\[
\mu(t) = e^{\int -\frac{a}{t} \, dt} = e^{-a \ln t} = t^{-a}
\]

Multiplying through the equation by \( \mu(t) = t^{-a} \):

\[
t^{-a} x'(t) - \frac{a}{t} t^{-a} x(t) = t^{-a}
\]

This simplifies to:

\[
\frac{d}{dt} \left( t^{-a} x(t) \right) = t^{-a}
\]

Integrating both sides:

\[
t^{-a} x(t) = \int t^{-a} \, dt
\]

The integral on the right-hand side depends on \( a \):

- For \( a \neq 1 \), we have:

\[
\int t^{-a} \, dt = \frac{t^{1-a}}{1-a}
\]

Thus, the solution is:

\[
t^{-a} x(t) = \frac{t^{1-a}}{1-a} + C_2
\]

Multiplying through by \( t^a \):

\[
x(t) = \frac{t}{1-a} + C_2 t^a
\]

- For \( a = 1 \), the equation becomes simpler, and we have:

\[
t^{-1} x'(t) = t^{-1}
\]

This integrates to:

\[
x(t) = \ln t + C_2
\]

### Step 3: Apply the initial condition \( x(1) = 1 \)

We now apply the initial condition \( x(1) = 1 \) to determine \( C_2 \).

- **Case 1**: For \( a \neq 1 \), we have:

\[
x(1) = \frac{1}{1-a} + C_2 = 1
\]

This gives:

\[
C_2 = 1 - \frac{1}{1-a} = \frac{-a}{1-a}
\]

Thus, the solution for \( a \neq 1 \) is:

\[
x(t) = \frac{t}{1-a} + \frac{-a}{1-a} t^a
\]

- **Case 2**: For \( a = 1 \), we have:

\[
x(1) = \ln 1 + C_2 = C_2 = 1
\]

Thus, the solution for \( a = 1 \) is:

\[
x(t) = \ln t + 1
\]

### Step 4: Maximum interval of existence

The general solution for \( a \neq 1 \) is valid for all \( t > 0 \), so the maximum interval of existence is \( (0, \infty) \).

For \( a = 1 \), the solution involves \( \ln t \), which is only defined for \( t > 0 \). Therefore, the interval of existence is also \( (0, \infty) \) in this case.

### Final Solution

- For \( a \neq 1 \), the solution is:

\[
x(t) = \frac{t}{1-a} - \frac{a t^a}{1-a}
\]

- For \( a = 1 \), the solution is:

\[
x(t) = \ln t + 1
\]

In both cases, the maximum interval of existence is \( (0, \infty) \).

---

Would you like more details on any part of this solution?

Here are 5 related questions you might find useful:
1. How does the method of integrating factors work in general?
2. What happens to the solution when \( a = 0 \)?
3. How do you classify and solve linear differential equations?
4. What is the significance of the initial condition \( x(1) = 1 \)?
5. Can we generalize this approach to systems of differential equations?

**Tip:** Always check for special values of parameters that could change the structure of the solution, like \( a = 1 \) in this problem.

Let a be some positive parameter. Solve the initial value problem, x(1) = 1, for x'(t) = (ax/t) + 1. What is the maximum interval of existence of this solution? Be careful: there exist(s) some special value(s) of the parameter for which the solution has a different form than for the other values. Find the general solution.

Solve the first-order linear differential equation x'(t) = (a x(t))/t + 1 with initial condition x(1) = 1. Explore the solution for different parameter values of a, using the method of integrating factors. Understand the maximum interval of existence and the unique behavior when a = 1.

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