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So I have f'(x)=-f(x)<=>f'(x)+f(x)=0, let h(x)=f'(x)+f(x)=0, I found two ways to approach this:

  1. e^x•f'(x)+e^x•f(x)=0<=>(e^x•f(x))'=0
  2. (f'(x)/f(x))+1=0<=>(ln(|f(x)|)+x)'=0 So h(x) is either 1) or 2), but I found that the 2) one is equal to ln(f(x)•e^x) so 1) and 2) definitely aren't the same initial functions and although I understand that them being equal is not necessary for their derivatives to be the same, the problem is that graphing the two initial functions it's clear to me that they have different derivatives, how is that possible? For f(x)=5x and other ones I tried the derivatives are different.enter image description here
Nick The Greek
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  • Please take some time to learn how to format math here, https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – it makes it so much easier to read & understand questions. – Gerry Myerson Nov 17 '23 at 12:05
  • I don't know what you mean by "initial functions", but $\log(e^xf(x))=\log(f(x))+x$, right? – Gerry Myerson Nov 17 '23 at 12:09
  • "For $f(x)=5x$ and other ones I tried (...)". The function $f(x)=5x$ does not satisfy the condition $f'+f = 0$. – PierreCarre Nov 17 '23 at 12:24

2 Answers2

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$$ (e^x f(x))'=0 $$ implies $e^xf(x)=c_1$ whereas $$ (\ln f(x)+x)'=0 $$ implies $\ln f(x)+x=c_2$. Assuming $\ln c_1=c_2$ they are the same, otherwise not.

R. J. Mathar
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Both formulations imply that $f(x) = c e^{-x}$, for some constant $c$. You can only fix the constant $c$ if you require some extra condition on $f$, for instance, its value in some prescribed point. If you decide that $f(0)=1$, what would be the expression for $f$ in each case?

PierreCarre
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