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

[mcastillo356]

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!

 

[DaveE]

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?

 

[mcastillo356]

Hi, PF, @DaveE

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.

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.

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!

 

[DaveE]

(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 $\displaystyle\int_4^x\,tdt = \frac{1}{2} [t^2]^x_4 = \frac{1}{2} (x^2 - 4^2) = \frac{1}{2} x^2 - 8$

ex. Am I in the path, closer to a right conclusion?

Yes.

 

[DaveE]

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

 

[mcastillo356]

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!

 

[DaveE]

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.

 

[DaveE]

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.