In Calc class, my teacher told us that the only solution to $y' = y$ is $y = ce^x$, with $c$ being a real number. I am having difficulty understanding the only part. Is there a proof of this? Or am I missing something obvious?

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    $\begingroup$ If $f$ is such a function, then $f(x)/e^x$ will have derivative zero. $\endgroup$ Jan 26, 2015 at 1:11
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    $\begingroup$ Is there a rigorous proof that a function whose derivative is everywhere zero must be a constant? Use mean value theorem? $\endgroup$
    – velut luna
    Jan 26, 2015 at 1:23

3 Answers 3


We have $y'=\frac{dy}{dx}=y$ where $y$ is a function of $x$. Therefore, by rearranging the equation,

\begin{align} \frac{dy}{y}=dx\,. \end{align}

Integrating with respect to $x$ on each side,

\begin{align} \int\frac{dy}{y}=\int dx\implies \ln(y)=x+C\,, \end{align}

where $C$ is an integration constant. Exponentiating both sides,

\begin{align} e^{\ln(y)}=e^{x+C}=e^Ce^x\,. \end{align}

As $e^C$ is just a constant, we will re-label it $c$. Also note $e^{\ln(y)}=y$. Thus $y=ce^x$.

There's probably a more formal way to see why $y=ce^x$, but I hope this explanation helps.


Expanding on Matthew's comment

Let $g(x)=f(x)/e^x$. Then, $g'(x)=\frac{f'(x)e^x-f(x)e^x}{e^{2x}}=0$ Therefore, $g(x)$ is a constant say $C$. Then, $f(x)=Ce^x$


The book Calculus by Tom Apostol defines ln(y) as $\ln(y) =\int \frac{1}{y} dy$. Using this definition the answer is clear as $\int \frac{dy}{y}= \int dx \implies \ln(y) = x+C $.

Following is a part of how $\ln(y)$ is defined. Consider an equation such that $f(xy)=f(x)+f(y)$ We can see that only solution valid for entire real line is $f(x)=0$ Now consider a function which is not defined at $0$.

The domain contains 1 Then $f(xy)=f(x)+f(y) \implies f(1)=2f(1)$ so $f(1)=0$


So $f(-x)=f(x)+f(-1)=f(x)$, so the function is even.

Now consider $ f'(xy)$ for a fixed $x$

By chain rule: $y f'(xy)=f'(x)$ as y is treated as a constant

When $x=1$ this gives us $y f'(y)=f'(1)$

So, $f'(y)=f'(1)\frac{1}{y}$

Here we may notice that $f'$ is monotonic and hence integrable on every closed interval not containing $0$.

$f(x)-f(c)=\int_c^xf'(t) dt=f'(1)\int_c^x\frac{1}{t}dt$

This equation holds if $x$ and $c$ are both graeter than $0$ or are both less than $0$. For a positive $x$ setting $c=1$

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

AS $f(x)$ is an even function we can have the definition $f(x)=f'(1)\int_1^{|x|}\frac{1}{t}dt$

Therefore solution to this equation is of form $f(x)=C\int_1^{|x|}\frac{1}{t}dt$ where $c \neq 0$

The definition of $\ln$ takes $C=1$ but any solution would be of the form given above i.e. of the form $C \ln(x)$.

In this way $\ln$ is defined as $\int_1^{|x|}\frac{1}{t}dt$ such that it is consistent with its definition as inverse of exponent. Note that the inverse of exponent definition is not easily explained for irrational values of $\ln (x)$ In this sense this function is a continuous function which is an extension of the usual definition.


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