1. ## Fredholm Intergral Equation

I need to find a general solution,

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

for the Fredholm intergral equation:

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

where $M$ is the intergral operator defined by

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

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

I think, that I have to rewrite it:

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

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

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

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:
$
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:
$f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$,
where $C_1=\int_0^{\infty} e^{-bt}f(t)dt=\int_0^{\infty} e^{-bt}(\gamma cost + \delta sin t + \lambda t^{a-1} C_1)dt$
Knowing $C_1$ you will find general solution by substituting it into the first equation in this message.

3. ## okay

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

I.e.

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

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

5. Well basically,

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

and subsitute it into

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

Then all I get is

$
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:
$f(x)= \gamma cos x + \delta sin x + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$
This equation has a solution:
$f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$
where $C_1=\int_0^{\infty} e^{-bt}f(t)dt$
Lets find $C_1$ in terms of $\lambda$:
$C_1=\int_0^{\infty} e^{-bt}f(t)dt=\int_0^{\infty} e^{-bt}(\gamma cost + \delta sin t + \lambda t^{a-1} C_1)dt = I_1 + I_2+ I_3$
$I_1=\gamma \int_0^{\infty} e^{-bt} cost dt = \frac{b \gamma}{b^2+1}$
$I_2=\delta \int_0^{\infty} e^{-bt} sint dt= \frac{\delta}{b^2+1}$
$I_3=-\lambda C_1 K$.
So:
$C_1=\frac{b \gamma + \delta}{b^2+1}-\lambda C_1 K$
$C_1=\frac{b \gamma + \delta}{(b^2+1)(1+\lambda K)}$
if $\lambda$ doesn't equal $-\frac{1}{K}$.
Now you should substitute $C_1$ into
$f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$

7. ## Thanks!

Thank you very much that was really helpful!!!

8. ## Fredholm Integral Equation

I need to find a general solution,

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

for the Fredholm intergral equation:

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

where $M$ is the intergral operator defined by

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

Where $a$ and $b$ are postive constants and $\gamma, \delta$ and $\lambda$∈ 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?

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

I would be really greatful if someone could help so that I could continue on the next part of the question please...