Prove the uniqueness of an integral equation

Question

$Question$

Prove that there is exactly one function $f,$ continuous on the positive real axis, such that

$$f(x) = 1+ \frac 1 x \int_1^x f(t) \, dt \text{ for all } x> 0$$ and find this function.

$Attempt$

Differentiating the about equation , we get

$$f'(x) = \frac{-1}{x^2}\int_1^x f(t) \, dt + \frac 1 x f(x)$$

Comparing with

$f'(x)+p(x) f( x) = q(x),$ we can see both $p(x)$ and $q(x)$ are continuous on $x> 0.$

Therefore $f(x)$ will be unique (uniqueness of $f(x)$ in $f'(x)+p(x) f(x) = q(x)$ can easily be proven if p and q are continuous in some interval I.)

$Doubt$

1. Is my proof correct?

2. How to find function $f(x)$?

Answer 1

Since $f(x)=1+\frac{1}{x}\int_1^xf(t) \, dt$, therefore we can use this in the equation $f^{'}(x)=\frac{-1}{x^2}\int_1^xf(t) \, dt+\frac{1}{x}f(x)$ to get \begin{align*} f^{'}(x) & =\frac{x(1-f(x))}{x^2}+\frac{1}{x}f(x)\\ & =\frac{x(1-f(x))}{x^2}+\frac{1}{x}f(x)\\ &=\frac{1}{x}. \end{align*} So $f(x)=\ln x+C$. Now we can use $f(1)=1$, to get $C$.

Answer 2

The equation is $xf(x)=x+\int_0^{x}f(t)\, dt$. Note that RHS is differentiable so $f$ is also differentiable. Differentiate to get $xf'(x)=1$ so $f(x)=\ln x+c$ and note that $f(1)=1$ from the given equation. So the solution is $f(x)=1+\ln x$. Your prrof is incorrect because you have not obtained the differential equation in the form $f'(x)+p(x)f(x)=q(x)$ (with $p$ and $q$ not depending of $f$).