To solve for the function $f(x)$ and the value $a$ such that:

$1 + \int_a^x \frac{f(t)}{t^3} \, dt = 5x^{-1}$

we'll follow these steps:

1. Differentiate Both Sides with Respect to $x$: By differentiating both sides, we aim to eliminate the integral and obtain an equation involving $f(x)$.

   Differentiating the left side:

   $\frac{d}{dx} \left(1 + \int_a^x \frac{f(t)}{t^3} \, dt\right) = \frac{f(x)}{x^3}$

   Differentiating the right side:

   $\frac{d}{dx}(5x^{-1}) = -5x^{-2}$

   Thus, we have:

   $\frac{f(x)}{x^3} = -5x^{-2}$

2. Solve for $f(x)$: Multiply both sides by $x^3$ to isolate $f(x)$:

   $f(x) = -5x$

3. Determine the Value of $a$: Now that we know $f(x) = -5x$, substitute back into the original equation to determine $a$.

   Substitute $f(x) = -5x$ into the integral:

   $1 + \int_a^x \frac{-5t}{t^3} \, dt = 5x^{-1}$

   Simplify the integrand:

   $1 + \int_a^x -\frac{5}{t^2} \, dt = 5x^{-1}$

   Now, integrate $-\frac{5}{t^2}$:

   $1 + \left[ \frac{5}{t} \right]_a^x = 5x^{-1}$

   Evaluating this gives:

   $1 + \left( \frac{5}{x} - \frac{5}{a} \right) = \frac{5}{x}$

   Simplify further:

   $1 - \frac{5}{a} = 0$

   Solving for $a$, we find:

   $a = 5$

So, the solution is:

$f(x) = -5x \quad \text{and} \quad a = 5$

Let me know if you'd like further clarification or details on any step.

Here are five related questions to expand your understanding:

1. How can we verify the solution by plugging $f(x)$ and $a$ back into the original equation?
2. What other functions $f(x)$ could satisfy similar integral equations?
3. How does differentiating under the integral sign help solve such problems?
4. Can this method be applied to other forms of integral equations?
5. What is the impact of the choice of $a$ in defining the integral?

Tip: When handling integral equations, differentiating both sides with respect to the variable of integration can simplify the problem by removing the integral.