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

[Phylosopher]

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]

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*}

 

[Phylosopher]

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]

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?

 

[Phylosopher]

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]

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

 

[PeroK]

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?

More generally:

https://math.berkeley.edu/~shinms/SP14-54/diffeq-thms.pdf

 

[Phylosopher]

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

I get your point. Thanks

More generally:

https://math.berkeley.edu/~shinms/SP14-54/diffeq-thms.pdf

I will give it a try and read it. Thanks

 

[FactChecker]

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.

 

[fresh_42]

Is this enough to say that it is the only solution?

You can use Picard-Lindelöf.

 

[member 587159]

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]

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/%7Eroy/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.