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I understand that assuming an analytic solution, we look at the Taylor series and arrive at a unique solution y = exp(x). However how do we know that there are no other non-analytic solutions? (Ideally with as little analysis machinery as possible)

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  • $\begingroup$ If we can assume $\int \frac 1x dx= \ln x + c$, it’s a straightforward task. $\endgroup$
    – Tavish
    May 2, 2020 at 11:55
  • $\begingroup$ See this answer math.stackexchange.com/a/1292586/72031 $\endgroup$
    – Paramanand Singh
    May 2, 2020 at 12:29
  • $\begingroup$ One can develop a full theory of exponential function based on the definition that $y=\exp(x) $ is the unique solution to $y'=y, y(0)=1$. This works out for complex variables also although a bit differently than explained in the linked answer. $\endgroup$
    – Paramanand Singh
    May 2, 2020 at 12:40

4 Answers 4

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Suppose $g'(x)=g(x)$ for all $x\in\mathbb R.$ $$ \underbrace{\frac d {dx}\,\frac{g(x)}{e^x} = \frac{e^x g'(x) - e^x g(x)}{e^{2x}}}_\text{quotient rule} = 0 \text{ for all } x\in\mathbb R. $$ Therefore $x\mapsto g(x)/e^x$ is constant on $\mathbb R.$

So $g(x) = \text{constant}\cdot e^x$ for $x\in\mathbb R.$

(The mean value theorem is tacitly used here, in that it is used in the proof that if the derivative of a function is $0$ everywhere in an interval, then the function is constant on that interval.)

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You can also make an argument that does not require you to know anything about the logarithm. suppose that $f$ and $g$ are two solutions to $y'=y$ with $f(x_0)=g(x_0)\neq 0$.

Then, by continuity, there exists $\delta$ such that $(x_0-\delta,x_0+\delta)\cap g^{-1}(\{0\})=\emptyset$. For $x\in (x_0-\delta,x_0+\delta),$ we have $$ \left(\frac{f}{g}\right)'(x)=\frac{f'g(x)-fg'(x)}{g^2(x)}=0, $$ So that $f=g$ on $(x_0-\delta,x_0+\delta)$. In fact, applying continuity, $f=g$ on $[x_0-\delta,x_0+\delta]$ and iterating this, we get that $f=g$ on the largest interval $I$ containing $x_0$ such that $g(y)\neq 0$ for all $y\in I$. However, picking $g(x)=\exp(x),$ we get that $I=\mathbb{R}$ and so, $f(x)=\exp(x)$.

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$\frac{dy}{dx} = y \Rightarrow \frac{dy}{y} = dx \Rightarrow \int \frac{dy}{y} = \int 1 \cdot dx \Rightarrow \ln|y| = x + c \Rightarrow y = C \cdot e^{x}$.

Now, because $y(0) = 1, C = 1$.

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  • $\begingroup$ This assumes, rather than proves, that $y$ is nowhere zero. $\endgroup$ May 2, 2020 at 12:08
  • $\begingroup$ This uses the FTC though $\endgroup$ May 2, 2020 at 12:11
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A proof using only the definitions of the derivative and the exponential function as limits, respectively $$ y'(x)=\lim_{h\rightarrow0}\left(\frac{y(x+h)-y(x)}{h}\right),~~(1) $$ and $$ e^x =\lim_{n\rightarrow \infty} \left(1+\frac{x}{n}\right)^{n}.~~(2)$$

From Eq.$~(2)$ we see that $e^x$ can also be written as $e^x =\lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^{nx}$.

Then, using Eq.$~(1)$ we convert the ODE $y'(x)=y(x)$ with initial conditions $y(0)=1$ to a recurrence equation $$y(x+h)-(1 + h) y(x)=0,$$ which can be solved (assuming a power-law $y\sim r^x$) to yield (after putting back the limit): $$y(x)=\lim_{h\rightarrow0}(1+h)^{x/h}=\exp(x),$$ where in the last part we used Eq.$~(2)$ and its variant.

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  • $\begingroup$ Why the downvote!? $\endgroup$
    – Hans Olo
    May 4, 2020 at 14:50

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