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

• I
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.

Homework Helper
Gold Member
Proof:
\begin{align*}
\frac{dy}{dx}&=y\\
\log y + C &= x + D\quad\ \ \textrm{[integrating both sides]}\\
y&=A e^x\quad\quad\ \ \textrm{[where $A=e^{D-C}$ is a constant]}
\end{align*}

FactChecker
Phylosopher
Proof:
\begin{align*}
\frac{dy}{dx}&=y\\
\log y + C &= x + D\quad\ \ \textrm{[integrating 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?

Homework Helper
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?

PeroK
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

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

Homework Helper
Gold Member
2021 Award
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

More generally:

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

I will give it a try and read it. Thanks

Gold Member
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.

Phylosopher
Mentor
2021 Award
Is this enough to say that it is the only solution?
You can use Picard-Lindelöf.

member 587159 and Phylosopher
Proof:
\begin{align*}
\frac{dy}{dx}&=y\\
\log y + C &= x + D\quad\ \ \textrm{[integrating 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)##

FactChecker