Solve Int. Eq.: Exponential Growth Diff. Eq.

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In summary: Attempt It is a shot in the dark. Many natural processes involve quantities that increase or decrease at a rate proportional to their size. All of these phenomena, can be modelled ##\displaystyle\frac{dy}{dt}=ky##, this is, the differential equation of exponential growth or decay.
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mcastillo356
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TL;DR Summary
I can't understand why the result of this integral equation is the differential equation of exponential growth
Hi, PF

There goes the solved example, the doubt, and the attempt:

Example 8 Solve the integral equation ##f(x)=2+\displaystyle\int_4^x\,f(t)dt##.

Solution Differentiate the integral equation ##f'(x)=3f(x)##, the DE for exponential growth, having solution ##f(x)=Ce^{3x}##. Now put ##x=4## into the integral equation to get ##f(4)=2##. Hence ##2=Ce^{12}##. So ##C=2e^{-12}##. Therefore, the integral equation has solution ##2e^{3x-12}##.

Doubt Why is the DE for exponential growth when we differentiate the integral equation to get ##f'(x)=3f(x)##?

Attempt It is a shot in the dark. Many natural processes involve quantities that increase or decrease at a rate proportional to their size. All of these phenomena, can be modelled ##\displaystyle\frac{dy}{dt}=ky##, this is, the differential equation of exponential growth or decay.

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Greetings!

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  • #2
mcastillo356 said:
TL;DR Summary: I can't understand why the result of this integral equation is the differential equation of exponential growth

Hi, PF

There goes the solved example, the doubt, and the attempt:

Example 8 Solve the integral equation ##f(x)=2+\displaystyle\int_4^x\,f(t)dt##.

Solution Differentiate the integral equation ##f'(x)=3f(x)##, the DE for exponential growth, having solution ##f(x)=Ce^{3x}##. Now put ##x=4## into the integral equation to get ##f(4)=2##. Hence ##2=Ce^{12}##. So ##C=2e^{-12}##. Therefore, the integral equation has solution ##2e^{3x-12}##.

Doubt Why is the DE for exponential growth when we differentiate the integral equation to get ##f'(x)=3f(x)##?

Attempt It is a shot in the dark. Many natural processes involve quantities that increase or decrease at a rate proportional to their size. All of these phenomena, can be modelled ##\displaystyle\frac{dy}{dt}=ky##, this is, the differential equation of exponential growth or decay.

PD: Post without preview
Greetings!

View attachment 329318
I'm somewhat confused by the details of your question. Like, why do you have the equation ##f(x)=2+\displaystyle\int_4^x\,f(t)dt## and ##f'(x)=3f(x)## related somehow (i.e. where did the "3" come from?).

But, let's take your exponential growth equation ##\displaystyle\frac{dy}{dt}=ky## and integrate both sides to get ##y(t)=C+k\displaystyle\int\,y(t)dt## where C is some constant. Does that look familiar?

What function is the same when you integrate it or take it's derivative?
 
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Hi, PF, @DaveE

DaveE said:
I'm somewhat confused by the details of your question. Like, why do you have the equation ##f(x)=2+\displaystyle\int_4^x\,f(t)dt## and ##f'(x)=3f(x)## related somehow (i.e. where did the "3" come from?).

Ok, I keep on dancing in the dark. Might it be that I must consider the ED like a given default information ?; or, how to say.... A fact of the problem that is not part of the solution; not a deduction, but a premise?. The circumstance that the integral equation involved, ##f(x)=2+3\displaystyle\int_4^x\,f(t)dt## is generic, not precise (the independent variable is ##t##) makes me think that, quoting Wikipedia, a differential equation relates one or more unknown functions and their derivatives.

DaveE said:
But, let's take your exponential growth equation ##\displaystyle\frac{dy}{dt}=ky## and integrate both sides to get ##y(t)=C+k\displaystyle\int\,y(t)dt## where C is some constant. Does that look familiar?

First of all, thanks for explaining from both points of view. I think it is very helpful.

DaveE said:
What function is the same when you integrate it or take it's derivative?

##e^x##. Am I in the path, closer to a right conclusion?

Greetings!
 
  • #4
mcastillo356 said:
(the independent variable is t) makes me think that, quoting Wikipedia, a differential equation relates one or more unknown functions and their derivatives.
##t## is what I would call a "dummy variable" it is simply describing what the integral applies to. ##\displaystyle\int_4^x\,f(t)dt## is a function of x, not ##t##. ##t## goes away in the evaluation of the integral.

For example consider the function ##\displaystyle\int_4^x\,tdt = \frac{1}{2} [t^2]^x_4 = \frac{1}{2} (x^2 - 4^2) = \frac{1}{2} x^2 - 8 ##

mcastillo356 said:
ex. Am I in the path, closer to a right conclusion?
Yes.
 
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  • #5
Suppose you guessed at a solution of ##f(x) = c + a⋅e^{bx}## where a, b, and c are unknown constants. If you substitute that into your equation can you deduce the values of a, b, and c?

Here is a more complicated example: https://en.universaldenker.org/lessons/1345
 
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DaveE said:
Suppose you guessed at a solution of ##f(x) = c + a⋅e^{bx}## where a, b, and c are unknown constants. If you substitute that into your equation can you deduce the values of a, b, and c?

Hi, PF, @DaveE, let me delay this exercise; the reason is that I want to set, fix concepts of the initial doubt.

First, here goes an outlined proof of ##\displaystyle\frac{dy}{dx}=ky\Leftrightarrow{y=Ce^{kx}}##

##\displaystyle\frac{dy}{y}=kdx##

##\displaystyle\int{\displaystyle\frac{dy}{y}}=\displaystyle\int{kdx}##

##\ln{|y|}=kx+C##

##\therefore{y=e^{kx+C}}##

It is not the topic I first proposed. And I still have one doubt now:

When talking about differentiating the integral equation, this is, ##f'(x)=3f(x)##, can I say it is inspired in ##\displaystyle\frac{dy}{dx}=ky##?

Greetings!
 
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  • #7
mcastillo356 said:
When talking about differentiating the integral equation, this is, f′(x)=3f(x), can I say it is inspired in dydx=ky?
yes, sure. k=3, a constant. k can be any constant in that equation.
 
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  • #8
mcastillo356 said:
First, here goes an outlined proof of ##\displaystyle\frac{dy}{dx}=ky\Leftrightarrow{y=Ce^{kx}}##

##\displaystyle\frac{dy}{y}=kdx##

##\displaystyle\int{\displaystyle\frac{dy}{y}}=\displaystyle\int{kdx}##

##\ln{|y|}=kx+C##

##\therefore{y=e^{kx+C}}##
Yes, excellent. The next, simple, step is to bring out the constant to match your original hypothesis. You've used C twice here in different places, so I'll redefine things:
##\displaystyle\frac{dy}{dx}=ky\Leftrightarrow{y=C_1e^{kx}}##

and the result:
##\therefore{y=e^{kx+C_2}} = e^{C_2}e^{kx}## so ##C_1 = e^{C_2}##, an arbitrary constant.

But this implies that ##C_1 > 0##. You have an alternate case, from the intermediate step you left out
##|y|=e^{kx+C_2}##, where ##-y=e^{kx+C_2}##, which will imply ##C_1 < 0##.

You can combine these all of this to allow any value of C_1. The ##C_1 = 0## case can be easily shown as a trivial solution.
 
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