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Is it possible to prove that $C_1e^x$ is the unique solution to $f'(x)=f(x)$?

I have tried to suppose there exists $g'=g$ and $g(x)\neq C_1e^x$. But I cannot find any contradiction by myself.

Any solutions or hints will be helpful. Thanks.

High GPA
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    Note that $C_2$ cannot take any arbitrary value to work in the solution here. – Tob Ernack Jul 10 '17 at 18:59
  • It isn't a solution to $f'=f$. If $f'=f$, think about $e^{-x}f(x)$. – Angina Seng Jul 10 '17 at 18:59
  • See this answer https://math.stackexchange.com/a/1292586/72031 – Paramanand Singh Jul 10 '17 at 19:02
  • It's a bad idea to rewrite your question in a way which makes a person's answer nonsensical. – Lee Mosher Jul 10 '17 at 20:01
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    Get in the habit of thinking of a differential equation as including a specification of the domain over which it is supposed to hold. For example: the equation $f'=f$ over $(-\infty, 0) \cup (0,\infty)$ has the solution $f(x)=-e^{-x} \text{ for } x<0\text{ and } f(x)=e^x\text{ for } x>0$ which does not have the form $C_1e^x$. –  Jul 11 '17 at 02:27

4 Answers4

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You can't have that $C_2$ there. But you know that $e^x$ is one solution. If $f$ is another, then $$\frac{\rm d}{{\rm d}x}\frac{f(x)}{e^x} = \frac{f'(x)e^x - f(x)e^x}{e^{2x}} = \frac{f(x)-f'(x)}{e^x} = 0,$$hence $f(x)/e^x = c$, and so $f(x) = ce^x$, for some $c \in \Bbb R$.

Ivo Terek
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Suppose $f$ is a solution. Then let $\phi(x) = e^{-x} f(x)$ and note that $\phi'(x) = 0$ and hence $\phi$ is a constant. Hence $f(x) = \phi(0) e^x$.

Hence the solution is unique (modulo the initial value).

copper.hat
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    This is the solution of IvoTerek. –  Jul 10 '17 at 19:15
  • @YvesDaoust: I think it predates Ivo & myself by a few decades... I make no claim to uniqueness (MSE is not about original mathematics), however, until you pointed it out, I didn't recognise Ivo's answer as similar to mine. – copper.hat Jul 10 '17 at 20:04
  • @YvesDaoust: Did you downvote because of this??? – copper.hat Jul 10 '17 at 20:05
  • No I didn't. It is useless to post twice the same. –  Jul 10 '17 at 21:08
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If $f(x)>0$, $f'(x)>0$ and $f$ is growing, hence remains positive. Conversely $f(x)<0$ remains negative. $f$ cannot vanish unless it is identically zero ($f(x)=0$).

Then

$$\frac{f'(x)}{f(x)}=\left(\log|f(x)|\right)'=1$$ and $$|f(x)|=e^{x+c}=Ce^x.$$

Alternatively, let $f,g$ be two solutions. Then

$$(\log f-\log g)'=\frac{f'}f-\frac{g'}g=0$$ so that $$\frac fg=C.$$

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The following proof is equivalent to the proofs of IvoTerek and copper.hat, but formulated a bit different.

Start with the differential equation $f' = f$. Rewrite it as $f' - f = 0$ and multiply with $e^{-x}$ (the integrating factor): $$e^{-x} f' - e^{-x} f = 0$$

The left hand side can now be written as the derivative of a product: $$e^{-x} f' - e^{-x} f = \left( e^{-x} f \right)'$$

Thus we have the differential equation $\left( e^{-x} f \right)' = 0$. Taking the antiderivative gives $e^{-x} f = C$ for some constant $C$. This gives $$f(x) = C e^x$$

md2perpe
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