The problem given in the image asks to find a function $f$ and a number $a$ such that:

$f(a) = \int_0^a \frac{f(t)}{t} \, dt = a^3 - 2$

The goal is to determine both the function $f(a)$ and the constant $a$ that satisfy this equation. Let's break it down:

Step 1: Understanding the Equation

You have a function $f(a)$ defined implicitly by an integral equation:

$f(a) = \int_0^a \frac{f(t)}{t} \, dt$

This integral needs to equal $a^3 - 2$ for some specific $a$.

Step 2: Differentiating Both Sides

Differentiate both sides of the equation with respect to $a$ to get rid of the integral:

[ \frac{d}{da}\left[f(a)\right] = \frac{d}{da}\left[\int_0^a \frac{f(t)}{t} , dt\right] ]

Using the Fundamental Theorem of Calculus, the right-hand side simplifies to:

$f'(a) = \frac{f(a)}{a}$

This is a differential equation.

Step 3: Solving the Differential Equation

The differential equation can be written as:

$\frac{f'(a)}{f(a)} = \frac{1}{a}$

Integrating both sides:

$\ln |f(a)| = \ln |a| + C$

Where $C$ is the constant of integration. Exponentiating both sides gives:

$f(a) = C a$

This indicates that $f(a)$ is a linear function of $a$, i.e., $f(a) = Ca$.

Step 4: Substitute Back into the Original Equation

Substitute $f(a) = Ca$ into the original integral equation:

$Ca = \int_0^a \frac{C t}{t} \, dt = \int_0^a C \, dt = Ca$

This is satisfied for any $C$.

Step 5: Using the Boundary Condition

Finally, use the given condition:

$f(a) = a^3 - 2$

Substitute $f(a) = Ca$ into this:

$Ca = a^3 - 2$

This simplifies to:

$C = a^2 - \frac{2}{a}$

So, $f(a) = a^3 - 2$ and this function is consistent with the integral form given that $a$ satisfies the equation $f(a) = a^3 - 2$.

Thus, the function $f(a) = a^3 - 2$ and the value $a = 1$ solve the equation.

Do you have any further questions, or would you like more detailed steps?

Related Questions:

1. What is the significance of the Fundamental Theorem of Calculus in solving integral equations?
2. How would the solution change if the boundary condition was different?
3. What is the general solution of the differential equation $f'(a) = \frac{f(a)}{a}$?
4. Can you explain why the constant $C$ doesn't affect the solution in this context?
5. What are some applications of this type of integral equation in physics or engineering?

Tip:

Always remember to verify your solution by substituting it back into the original equation to check for consistency.