I Is the exponential function, the only function where y'=y?

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Hello,


I was wondering. Is the exponential function, the only function where ##y'=y##.

I know we can write an infinite amount of functions just by multiplying ##e^{x}## by a constant. This is not my point.

Lets say in general, is there another function other than ##y(x)=ae^{x}## (##a## is a constant), where ##\frac{dy}{dx}=y##.


I would really appreciate it if we can work a proof.
 

andrewkirk

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Proof:
\begin{align*}
\frac{dy}{dx}&=y\\
\frac{dy}y&=dx\quad\quad\quad\textrm{[separation of variables]}\\
\log y + C &= x + D\quad\ \ \textrm{[integrating both sides]}\\
y&= e^xe^{D-C}\quad\textrm{[exponentiating both sides]}\\
y&=A e^x\quad\quad\ \ \textrm{[where $A=e^{D-C}$ is a constant]}
\end{align*}
 
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Proof:
\begin{align*}
\frac{dy}{dx}&=y\\
\frac{dy}y&=dx\quad\quad\quad\textrm{[separation of variables]}\\
\log y + C &= x + D\quad\ \ \textrm{[integrating both sides]}\\
y&= e^xe^{D-C}\quad\textrm{[exponentiating both sides]}\\
y&=A e^x\quad\quad\ \ \textrm{[where $A=e^{D-C}$ is a constant]}
\end{align*}
Exactly. But I was thinking. Is this enough to say that it is the only solution?

I know this sounds hilarious, but when I studied calculus no one pointed out that the solution is unique. Is it?
 

lurflurf

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we have
##\dfrac{d}{dx}y=y##
let us rewrite it as
##\dfrac{d}{dx}e^{-x}y=0##
now consider
##\dfrac{d}{dx}u=0##
what can we say about u?
 
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we have
##\dfrac{d}{dx}y=y##
let us rewrite it as
##\dfrac{d}{dx}e^{-x}y=0##
now consider
##\dfrac{d}{dx}u=0##
what can we say about u?
Constant w.r.t x
 

lurflurf

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^then the only solutions are
##e^{-x}y=C##
as desired
 

PeroK

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Exactly. But I was thinking. Is this enough to say that it is the only solution?
Step-by-step, there are no alternatives except for the choice of C and D.
 

Math_QED

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Proof:
\begin{align*}
\frac{dy}{dx}&=y\\
\frac{dy}y&=dx\quad\quad\quad\textrm{[separation of variables]}\\
\log y + C &= x + D\quad\ \ \textrm{[integrating both sides]}\\
y&= e^xe^{D-C}\quad\textrm{[exponentiating both sides]}\\
y&=A e^x\quad\quad\ \ \textrm{[where $A=e^{D-C}$ is a constant]}
\end{align*}
You divided by ##y##, which is not allowed when ##y = 0##.

This happens to be a solution, which can be included if you take ##y = A\exp(x)## with ##A \geq 0## as general solution (instead of ##A > 0)##
 

mathwonk

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to expand on lurflurf's solution, use the quotient rule to show that if y'=y, then (y/e^x)' = 0, hence y/e^x = constant.

using this a lemma, one can deduce that the kernel of the linear constant coefficient operator P(D), where Df = f' and P is a polynomial of degree n, has dimension n. see my web notes on linear algebra, last section of chapter 5, roughly page 119:

http://alpha.math.uga.edu/~roy/laprimexp.pdf

briefly, the argument above gives the kernel of the operator (D-1) which generalizes to that of (D-c) and then by factoring a polynomial P(D) into products of linear factors of form (D-c), we get the result.
 
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