How to Solve an Integral Equation Involving Exponential Functions?

[Sunny29]

Please anyone can help solve this integral equation
e^t+e^t ∫ (t, 0 ) e^(-τ) x f(τ) dτ

 

[DrClaude]

What have you tried? And what is f(τ)?

 

[Delta2]

Well one can never be sure but I suppose f(t) is the unknown function we wish to find. But first things first, is the equation we trying to solve as follows
$e^t+e^t\int_0^t {e^{-\tau}f(\tau)d\tau}=0$?

 

[Sunny29]

What have you tried? And what is f(τ)?

Sorry, i haven't tried because i am not understanding the question completely.

 

[Sunny29]

Well one can never be sure but I suppose f(t) is the unknown function we wish to find. But first things first, is the equation we trying to solve as follows
$e^t+e^t\int_0^t {e^{-\tau}f(\tau)d\tau}=0$?

No, it is not equal to 0.
The question is exactly like this
"Solve the Integral Equation
f(t)= same as you have written

 

[Delta2]

Ok so the equation is
$f(t)=e^t+e^t\int_0^t {e^{-\tau}f(\tau)d\tau}$.

First do some algebraic operations and write the equation in the form $\int_0^t {e^{-\tau}f(\tau)d\tau}=...$
Now what operation can you apply at that form (hint: then plan is to convert the integral equation to an ordinary differential equation, there might be other ways to do this problem)

 

[eys_physics]

It is easy to show that the equation can be transformed to the form
$$g(t)=1+\int_0^t g(\tau)d\tau$$,
where
$$g(t)=e^{-t}f(t)$$.

The equation can as mentioned before be transformed to a differential equation. A more simple way is however to solve it directly by iteration, i.e.
$$g^{(n)}(t)=1+\int_0^t g^{(n-1)}(\tau)d\tau$$.
The iteration is started by using $g^{(0)}=1$ and then computing $g^{(1)}$, $g^{(2)}$, etc. After a few iterations it should be obvious what $g(x)$ is.

 

[Sunny29]

It is easy to show that the equation can be transformed to the form
$$g(t)=1+\int_0^t g(\tau)d\tau$$,
where
$$g(t)=e^{-t}f(t)$$.

The equation can as mentioned before be transformed to a differential equation. A more simple way is however to solve it directly by iteration, i.e.
$$g^{(n)}(t)=1+\int_0^t g^{(n-1)}(\tau)d\tau$$.
The iteration is started by using $g^{(0)}=1$ and then computing $g^{(1)}$, $g^{(2)}$, etc. After a few iterations it should be obvious what $g(x)$ is.

Dear,
Could you please solve the complete answer step by step? I am having difficulty in understanding.
Thanks..!

 

[eys_physics]

As, I understand it is against the rules at this forum to give a complete answer to the problem. You need to show the work you have done.
But, if you tell me which step you don't understand I will be happy to help. So, which step do you don't understand?

 

[Sunny29]

As, I understand it is against the rules at this forum to give a complete answer to the problem. You need to show the work you have done.
But, if you tell me which step you don't understand I will be happy to help. So, which step do you don't understand?

okay no problem. you just tell me how you write the first step and how you put the 1.?

 

[eys_physics]

okay no problem. you just tell me how you write the first step and how you put the 1.?

You have the equation
$$f(t)=e^t+e^t\int_0^{t}e^{-\tau}f(\tau)d\tau\quad (1)$$

Do you understand how to go from Eq. (1) to
$$g(t)=1+\int_0^{t}g(\tau)d\tau\quad (2) $$,
where $g(t)=e^{-t}f(t)$ ?

 

[Sunny29]

You have the equation
$$f(t)=e^t+e^t\int_0^{t}e^{-\tau}f(\tau)d\tau\quad (1)$$

Do you understand how to go from Eq. (1) to
$$g(t)=1+\int_0^{t}g(\tau)d\tau\quad (2) $$,
where $g(t)=e^{-t}f(t)$ ?

No, i can't understand how to go from Eq 1. to Eq 2.
Also, i am bit confusing between "Tau" and "t"

 

[eys_physics]

Multiply both sides of Eq. (1) by $e^{-t}$ and then simplify. What do you get?

Both $t$ and $\tau$ are variable. For each value of $t$ we have one a left-hand side an integration from $0$ to $t$. Therefore, to not mess things up we need to introduce an integration variable called $\tau$. Obviously, the name of the variable is arbitrary.

 

[Sunny29]

Multiply both sides of Eq. (1) by $e^{-t}$ and then simplify. What do you get?

Both $t$ and $\tau$ are variable. For each value of $t$ we have one a left-hand side an integration from $0$ to $t$. Therefore, to not mess things up we need to introduce an integration variable called $\tau$. Obviously, the name of the variable is arbitrary.

yes i have got. now what is after the iteration step? g(x)=?

 

[eys_physics]

yes i have got. now what is after the iteration step? g(x)=?

As, I understand you now got Eq. (2). Now, we want to solve this by iteration.
On the left-hand side we start by a "guess" for $g(\tau)$. As, the inhomogeneous term is 1. We start by putting $g(\tau)=1$ on the left-hand side.
You now compute a better estimate $g(t)$ from the equation. This should give you $g(t)=1+t$.
Now, in the next step we put $g(\tau)=1+\tau$ on the left-hand side and calculate a new $g(t)$. This procedure should lead to
$$g(t)=1+t+t^2/2+t^3/6+...$$
Do you get this?

 

[Sunny29]

As, I understand you now got Eq. (2). Now, we want to solve this by iteration.
On the left-hand side we start by a "guess" for $g(\tau)$. As, the inhomogeneous term is 1. We start by putting $g(\tau)=1$ on the left-hand side.
You now compute a better estimate $g(t)$ from the equation. This should give you $g(t)=1+t$.
Now, in the next step we put $g(\tau)=1+\tau$ on the left-hand side and calculate a new $g(t)$. This procedure should lead to
$$g(t)=1+t+t^2/2+t^3/6+...$$
Do you get this?

Yes, for sure i understand this also..
Does it ends here or something remaining?

 

[eys_physics]

Now, if you recall the Maclaurin expansion of $e^t$, see https://en.wikipedia.org/wiki/Taylor_series , it follows
$g(t)=e^t$.
From this we obtain $f(t)$.

 

[Sunny29]

Now, if you recall the Maclaurin expansion of $e^t$, see https://en.wikipedia.org/wiki/Taylor_series , it follows
$g(t)=e^t$.
From this we obtain $f(t)$.

Yes i have solved it. And thank you very much for your help.

 

[eys_physics]

You are welcome.