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I'm given $V = \{f(x) : R \rightarrow R : f'(x) = 2f(x), \forall x \in R\}$.

The only function which obeys the rule is $e^{2 x}$ or multiples of it. How can I prove that there is no other?

Git Gud
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0 kelvin
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2 Answers2

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Let $f(x)$ be an solution to the differential equation $y'=2y$. Then use the product rule to find the derivative of $g(x)=f(x)\cdot e^{-2x}$. If the derivative is zero, then $g(x)$ is constant, and that would say something about $f(x)$.

Rory Daulton
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  • OK. But I'm trying to figure a simple way that doesn't involve differential equations. – 0 kelvin Mar 08 '15 at 20:48
  • @0kelvin: What do you mean? There is a differential equation right there in your question. Do you mean you do not want to solve a differential equation? That was the point of my question, to avoid exactly that. The fact that a function with a zero derivative is a constant is a well-known theorem, coming directly from the Mean Value theorem. – Rory Daulton Mar 08 '15 at 20:51
  • @0kelvin: I meant "that was the point of my answer". – Rory Daulton Mar 08 '15 at 20:57
  • I meant that this exercise has a differential equation but it ommits the term, meaning that I'm assuming that it's not a prerequisite to know differential equations to solve it. – 0 kelvin Mar 08 '15 at 21:51
  • @0kelvin: Then my answer works. Start it with "Let $f(x)$ be a function such that $f\in V$" or more simply "Let $f\in V$." The knowledge needed is the product rule, the derivative of $e^{-2x}$, and the Mean Value Theorem. There is no mention of or solution of a differential equation. – Rory Daulton Mar 08 '15 at 22:46
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The best idea would be to just solve it.

$y'=\frac{dy}{dx}=2y$ implies that $\int \frac{1}{y}dy = \int 2dx$, and so $\ln(y)=2x+c$, or $y=c_1e^{2x}$

From the existence and uniqueness theorem, we know that the space of solutions for this ode is spanned only by the solution we found earlier.

Oria Gruber
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  • Your solution method assumes that $y\ne 0$, which is not necessarily true. Also, there should be an absolute value inside the logarithm. You could handle those cases and end up with your answer. Since the OP asked for a proof, those cases probably should be explicitly handled. – Rory Daulton Mar 08 '15 at 20:49