1. ## Fredholm Intergral Equation

I need to find a general solution,

$\displaystyle f$ ∈ $\displaystyle E [0, \infty)$

for the Fredholm intergral equation:

$\displaystyle f(x) = \gamma cos (x) + \delta sin (x) + \lambda (Mf)(x)$

where $\displaystyle M$ is the intergral operator defined by

$\displaystyle (Mf)(x):= \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$

Where $\displaystyle a$ and $\displaystyle b$ are postive constants and $\displaystyle \gamma, \delta$ and $\displaystyle \lambda$∈ ℝ constants.

I think, that I have to rewrite it:

$\displaystyle f(x)= \gamma cos (x) + \delta sin (x) + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$

And then assuming that λ is a fixed unknown scalar, and so can be incorporated into the equation as follows:

$\displaystyle f(x)= \gamma cos (x) + \delta sin (x) +\int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$

I'm not sure if this is right, and I don't really understand what to do next, can someone help please?

2. IMHO
The next step after this:
$\displaystyle f(x)= \gamma cos x + \delta sin x + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$
is to say: if equation invloved has a solution, its solution can be represented in the following way:
$\displaystyle f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$,
where $\displaystyle C_1=\int_0^{\infty} e^{-bt}f(t)dt=\int_0^{\infty} e^{-bt}(\gamma cost + \delta sin t + \lambda t^{a-1} C_1)dt$
Knowing $\displaystyle C_1$ you will find general solution by substituting it into the first equation in this message.

3. ## okay

Thanks, for that, but if I substitute $\displaystyle C_1$ into the first equation, then won't I just be back where I started?

I.e.

Just with $\displaystyle x$ replaced with $\displaystyle t$'s

4. Did you check this?
Note that you have equation for $\displaystyle C_1$ where this stuff takes part in both sides of the equation.

5. Well basically,

if I let $\displaystyle C_1=\int_0^{\infty} e^{-bt}f(t)dt$

and subsitute it into

$\displaystyle f(x)= \gamma cos x + \delta sin x + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$

Then all I get is

$\displaystyle f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$

And keep going round in circles.

Maybe at this point I need to solve the integral, but I am not sure how to do this?

6. No. Look at this:
You have:
$\displaystyle f(x)= \gamma cos x + \delta sin x + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$
This equation has a solution:
$\displaystyle f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$
where $\displaystyle C_1=\int_0^{\infty} e^{-bt}f(t)dt$
Lets find $\displaystyle C_1$ in terms of $\displaystyle \lambda$:
$\displaystyle C_1=\int_0^{\infty} e^{-bt}f(t)dt=\int_0^{\infty} e^{-bt}(\gamma cost + \delta sin t + \lambda t^{a-1} C_1)dt = I_1 + I_2+ I_3$
$\displaystyle I_1=\gamma \int_0^{\infty} e^{-bt} cost dt = \frac{b \gamma}{b^2+1}$
$\displaystyle I_2=\delta \int_0^{\infty} e^{-bt} sint dt= \frac{\delta}{b^2+1}$
$\displaystyle I_3=-\lambda C_1 K$.
So:
$\displaystyle C_1=\frac{b \gamma + \delta}{b^2+1}-\lambda C_1 K$
$\displaystyle C_1=\frac{b \gamma + \delta}{(b^2+1)(1+\lambda K)}$
if $\displaystyle \lambda$ doesn't equal $\displaystyle -\frac{1}{K}$.
Now you should substitute $\displaystyle C_1$ into
$\displaystyle f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$

7. ## Thanks!

Thank you very much that was really helpful!!!

8. ## Fredholm Integral Equation

I need to find a general solution,

$\displaystyle f$ ∈ $\displaystyle E [0, \infty)$

for the Fredholm intergral equation:

$\displaystyle f(x) = \gamma cos (x) + \delta sin (x) + \lambda (Mf)(x)$

where $\displaystyle M$ is the intergral operator defined by

$\displaystyle (Mf)(x):= \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$

Where $\displaystyle a$ and $\displaystyle b$ are postive constants and $\displaystyle \gamma, \delta$ and $\displaystyle \lambda$∈ ℝ constants.

Note that the algebra can be reduced using the Gamma function and standard Laplace Transforms.

I have started a solution to this, which is attached, can anyone tell me if this is all I have to do to get the general solution or is there more, and if so can anyone advise what I do next please?

9. ## My Solution??? Is it correct?

$\displaystyle f(x) = \gamma cos (x) + \delta sin (x) + \frac{\lambda x^{a-1}(b \gamma + \delta)}{(b^2+1)(1+\lambda K)}$