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 $(0, \infty)$.

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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.