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Find the function \(f(x)\) that satisfies:

\(f(x) = 1 - \int_0^x [f(t)]^2 \ dt\),

for all \(x \geq 0.\)

 

Thanks!

 Dec 24, 2021
 #1
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$f(x) = \frac{x}{(x + 1)^2}$.

 Dec 24, 2021
 #2
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Answer #1 is not correct.

Best is to differentiate wrt x, turning the integral equation into a differential equation.

\(\displaystyle \frac{d}{dx}f(x)= -\{1.[f(x)]^{2} - 0.[f(0)]^{2} +\int^{x}_{0}\frac{\partial}{\partial x}[f(t)]^{2}dt , \)

\(\displaystyle \frac{d}{dx}f(x)= - [f(x)]^{2}.\)

Now integrate that.

There will be a constant of integration, so substitute back into the original equation to determine its value.

 Dec 25, 2021
 #3
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Thanks Tiggsy,

Unfortunately, I do not understand how you have differentiated the RHS at all.   sad

Maybe you could explain a bit more please?

Melody  Dec 27, 2021
edited by Melody  Dec 27, 2021
 #4
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Hi Melody, the RHS comes from a standard result usually referred to as ' differentiation under the integral sign '.

For an indefinite integral the result takes the form

\(\displaystyle \frac{\partial}{\partial \alpha}\int f(\alpha,x)dx = \int\frac{\partial}{\partial \alpha}f(\alpha,x)dx .\)

It says that the order of the integral and the differentiation thingy are interchangeable, alpha is a parameter.

That's not always true, but it will be true for all of the functions you or I are ever going to meet.

As a simple example, since

\(\displaystyle \int e^{\alpha x} dx = \frac{1}{\alpha}e^{\alpha x}, \text{then}\\ \frac{\partial}{\partial \alpha} \int e^{\alpha x}dx=\int\frac{\partial}{\partial \alpha}e^{\alpha x} dx= \int xe^{\alpha x}dx =\frac{\partial}{\partial \alpha}(\frac{1}{\alpha}e^{\alpha x})= (\frac{x}{\alpha}-\frac{1}{\alpha^{2}})e^{\alpha x}.\)

(You can check the result by integrating by parts).

The result also works for a definite integral, providing that the limits of integration are constants.

If one or both of the limits of integration are functions of the parameter then the result takes the form

\(\displaystyle \frac{\partial}{\partial \alpha} \int^{b}_{a}f(\alpha,x)dx =\int^{b}_{a}\frac{\partial}{\partial \alpha}f(\alpha,x)dx+ \frac{db}{d \alpha}f(\alpha,b)-\frac{da}{d \alpha}f(\alpha,a).\)

For the example in the question, the parameter is x and the integration is wrt t, so we have (for the integral),

\(\displaystyle \frac{\partial}{\partial x}\int^{x}_{0}[f(t)]^{2}dt= \int^{x}_{0}\frac{\partial}{\partial x}[f(t)]^{2} dt +\frac{d}{d x}(x).[f(x)]^{2}-\frac{d}{dx}(0).[f(0)]^{2} \\ =[f(x)]^{2}.\)

Differentiate a function of t partially wrt x and you get zero, differentiate zero wrt x and you get zero so the first and third terms disappear.

 

***************************************************************************************************************************************

I hope that Christmas has gone well for you, and my very best wishes for the new year.

Tiggsy

 Dec 30, 2021
 #5
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Thanks very much Tiggsy

I don't remember that formula but I enjoyed working through it.

 

I hope you also had a great Christmas and that the year to come is excellent for you   laugh

Melody  Jan 1, 2022
 #6
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I followed what you did so far Tiggsy  BUT I still don't know how to finish it.

 

Maybe I have brain freeze.   sad

 

 

\(f(x) = 1 - \int_0^x [f(t)]^2 \ dt\\ \)

\(\displaystyle \frac{d}{dx}f(x)= - [f(x)]^{2}\\ \int\displaystyle \frac{d}{dx}f(x)dx= -\int [f(x)]^{2}dx\\ f(x)= -\displaystyle \int [f(x)]^{2}dx\\ \)

 

???

 Jan 2, 2022
 #7
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Hi Melody

Second line down, let y = f(x) and you have the equation   

\(\displaystyle \frac{dy}{dx}=-y^{2}, \text{ from which } \int \frac{dy}{y^{2}}=-\int dx. \)

Solve that for y  ( = f(x) ) and substitute the result back into the original integral equation to determine the value of the constant of integration.

You should finish up with the result f(x) = 1/(x + 1).

 

Tiggsy

 Jan 2, 2022

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