+0

# Find the function f(x) that satisfies...

0
76
7
+45

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
0

$f(x) = \frac{x}{(x + 1)^2}$.

Dec 24, 2021
#2
+199
+1

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
+115924
0

Thanks Tiggsy,

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

Maybe you could explain a bit more please?

Melody  Dec 27, 2021
edited by Melody  Dec 27, 2021
#4
+1

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
+115924
0

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

Melody  Jan 1, 2022
#6
+115924
0

I followed what you did so far Tiggsy  BUT I still don't know how to finish it.

Maybe I have brain freeze.

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

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

???

Jan 2, 2022
#7
0

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