What function f has the following property f(x)=1+1/x * integral(from 1 to x)f(t)dt ?

Question

How can i prove that f is unique?
f is defined on (0,infinity) and f is continuous

Answer 1

f(x) = 1 + 1/x [1, x]∫f(t) dt

Let F(x) = [1, x]͏∫f(t) dt. Since f is continuous, by the second fundamental theorem of calculus, we have F'(x) = f(x). So:

F'(x) = 1+1/x F(x)
F'(x) - 1/x F(x) = 1

We need an integrating factor here, which is e^(∫-1/x dx) = e^(-ln x) = 1/x. Multiplying both sides by 1/x:

1/x F'(x) - 1/x² F(x) = 1/x

Using the product rule and integrating:

d/dx (1/x F(x)) = 1/x
1/x F(x) = ∫1/x dx = ln x + C

Solving for F and differentiating to obtain f:

F(x) = x ln x + Cx
f(x) = F'(x) = 1 + ln x + C

Of course, this only shows that if f(x) = 1 + 1/x [1, x]∫f(t) dt, then f(x) = 1 + ln x + C. We also wish to show the reverse. Let us plug our solution into the original problem:

1 + 1/x [1, x]∫f(t) dt
1 + 1/x [1, x]∫1 + ln x + C dt
1 + 1/x (t ln t + Ct|[x, 1])
1 + 1/x (x ln x + Cx - C)
1 + ln x + C - C/x
f(x) - C/x

Note that this is equal to f(x) if and only if C=0. So f(x) = 1 + ln x is the unique solution to this problem.

Answer 2

in actuality what you are going to do is to replace x for each x interior the function and back with an identical function replace x-a million for each x interior the function. in the event that they equate then that could be the function. so which you may attempt it the 5 possibilities. (no longer complication-free artwork :) occasion: decision a) f(x)= a million-x; (yawn) f(a million-x) = a million - (a million-x) = +x actual they do no longer seem to be equivalent. decision b) f(x) = a million - x^2 f(a million-x) = a million - (a million-x)^2 = a million - a million +2x -x^2 = 2x -x^2 No sturdy. decision c) f(x) = x^2 - (a million-x)^2 f(a million-x) = (a million-x)^2 - (a million-(a million-x))^2 = (a million-x)^2 - x^2 No sturdy decision d) f(x) = x^2(a million-x)^2 f(a million-x) = (a million-x)^2 (a million-(a million-x))^2 = (a million-x)^2 x^2 This could be it. Yawn. The final would be undefined at x=a million. So throw it out the window proper away.

Answer 3

you have
x(f(x) -1) = integral...
you take derivative and use fundamental theorem of calculus:

(x(f(x)-1)) ' = f(x)

and now you solve this first degree differential equation

actually f(x) will simplify and is easier.

Answer 4

1 + 1/x (x ln x + Cx - C)
1 + ln x + C - C/x
f(x) - C/x