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So I need to prove that any function that fulfills Equation 1 also fulfills Equation 2 for an arbitrary small value of $k.$

Equation 1: $f'(x)=k\frac{f(x)}{f(x)+1-k}$

Equation 2: $f(n+1)-f(n)=k\frac{f(n)}{f(n)+1-k}$

What I have tried: I have been trying to integrate Equation 1 with a definite integral from $n$ to $n + 1$ and I'm currently stuck trying to figure out what the integral of $\frac{1}{f(x)+1}$ is.

Can someone let me know if I'm on the right track and what the integral of $\frac{1}{f(x)+1}$ is??

Thanks guys!

Dying at an increasing rate
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    $$ \frac{df}{dx} = k \frac{f}{f+1-k} $$ $$ \frac{(f+1-k) ,df} f = dx $$ $$ \left( 1 + \frac{1-k} f \right) df = dx $$ etc. – Michael Hardy Dec 09 '17 at 23:33
  • maybe try the mean value theorem? You can't integrate $f$ without knowing what it is, but the left hand side of the second equation is a difference quotient for $f$. – Sort of Damocles Dec 09 '17 at 23:34
  • This is all the information I was given for the problem (not hw, don't worry). I have no $f(x)$ given. How do I prove this problem though? I know I cannot use any direct proofs or induction so I don't know what this problem wants as far as proving it. – Dying at an increasing rate Dec 09 '17 at 23:43
  • This separation of variables readily gives you $x$ as a function of $f,$ but the inverse of that seems not as clearcut. However, it is not yet clear to me that one must solve this differential equation in order to answer the question. – Michael Hardy Dec 10 '17 at 03:06
  • @MichaelHardy: I need to prove that for $n\leq x\lt n+1$, why $f(x)=f(n)$ and integrate $f'(x)=k_\frac{f(n)}{f(n)+1-k}$ from $n$ to $n+1$, with the right side as a constant. – Dying at an increasing rate Dec 10 '17 at 18:18

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