# A Solve the Integral Equation

1. Nov 29, 2016

### Sunny29

• Member warned that homework questions must be posted in the Homework sections
Please anyone can help solve this integral equation
e^t+e^t ∫ (t, 0 ) e^(-τ) x f(τ) dτ

2. Nov 29, 2016

### Staff: Mentor

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

3. Nov 29, 2016

### 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
$e^t+e^t\int_0^t {e^{-\tau}f(\tau)d\tau}=0$?

4. Dec 1, 2016

### Sunny29

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

5. Dec 1, 2016

### Sunny29

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

6. Dec 1, 2016

### 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)

7. Dec 1, 2016

### 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.

8. Dec 3, 2016

### Sunny29

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

9. Dec 3, 2016

### 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?

10. Dec 3, 2016

### Sunny29

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

11. Dec 3, 2016

### eys_physics

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 $g(t)=e^{-t}f(t)$ ?

12. Dec 4, 2016

### Sunny29

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

13. Dec 4, 2016

### 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.

14. Dec 4, 2016

### Sunny29

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

15. Dec 4, 2016

### eys_physics

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?

16. Dec 4, 2016

### Sunny29

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

17. Dec 4, 2016

### eys_physics

18. Dec 5, 2016

### Sunny29

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

19. Dec 5, 2016

### eys_physics

You are welcome.