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

In summary, the conversation discusses the uniqueness of the solution for the differential equation ##y'=y## and how it can be proven using the Picard-Lindelöf theorem. The summary also touches on the general solution for the equation and how it can be expanded using the quotient rule. Finally, the summary mentions how this can be applied to prove the dimension of the kernel of a linear constant coefficient operator.
  • #1
Phylosopher
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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.
 
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  • #2
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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  • #3
andrewkirk said:
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?
 
  • #4
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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  • #5
lurflurf said:
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
 
  • #6
^then the only solutions are
##e^{-x}y=C##
as desired
 
  • #7
Phylosopher said:
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
 
  • #8
lurflurf said:
^then the only solutions are
##e^{-x}y=C##
as desired

I get your point. Thanks

PeroK said:
More generally:

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

I will give it a try and read it. Thanks
 
  • #9
Phylosopher said:
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.
 
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  • #10
Phylosopher said:
Is this enough to say that it is the only solution?
You can use Picard-Lindelöf. :wink:
 
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  • #11
andrewkirk said:
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)##
 
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  • #12
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.
 
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1. Is the exponential function the only function where y'=y?

No, there are other functions where y'=y, such as the sine and cosine functions. However, the exponential function is unique in that it is the only function where the rate of change is proportional to the current value.

2. What makes the exponential function special?

The exponential function is special because it is the only function where the rate of change is proportional to the current value. This property makes it useful in modeling growth and decay processes in many fields of science.

3. Can the exponential function be used to model all types of growth?

No, the exponential function can only model exponential growth, which is growth that increases at an increasing rate. Other types of growth, such as linear and logarithmic, require different mathematical functions.

4. What is the relationship between the exponential function and natural logarithms?

The exponential function and natural logarithms are inverse functions of each other. This means that when the exponential function is applied to a number, the natural logarithm of that number will give the original value. Similarly, when the natural logarithm is applied to a number, the exponential function will give the original value.

5. Can the exponential function be used in real-world applications?

Yes, the exponential function is used in many real-world applications, such as population growth, compound interest, and radioactive decay. It is a fundamental tool in fields such as economics, biology, and physics.

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