# Solving Integral Equation with Laplace Transform

1. Jul 17, 2010

### MisterMan

1. The problem statement, all variables and given/known data

$$y(x) = e^{3x} + \int_0^{x}\hspace{1mm}y(t)\hspace{1mm}e^{x-t}\hspace{1mm}dt$$

2. Relevant equations

$$(f*g)(x) = \int_0^{x}\hspace{1mm}y(t)\hspace{1mm}g(x-t)\hspace{1mm}dt$$

$$L(f*g;s) = L(f;s)L(g;s)$$

I will use $$Y(s)$$ to denote $$L(y;s)$$

3. The attempt at a solution

I tried to solve this like all the other problems I have encountered so far, I took the laplace transform of both sides, giving:

$$Y(s) = s-3 + (s-1)Y(s)$$

Which gave me:

$$Y(s) = -\frac{s-3}{s-2}$$

However, this doesn't work out, as far as I'm aware there is no inversion for 1 ( I have only dealt with the standard Laplace transforms and can only invert back to them ).

The book I have gives:

$$Y(s) = \frac{s-1}{(s-2)(s-3)}$$

But I'm not sure how this was calculated, I'll appreciate any hints and pointing out where I have gone wrong, thanks.

2. Jul 17, 2010

### samalkhaiat

$$Y(s) = \frac{1}{s-3} + \frac{Y(s)}{s-1}$$

You can also differentiate with respect to x to get

$$y^{'}(x) = 2 e^{3x} + 2y(x)$$

now take your transform and use

$$L(y^{'}) = sY(s) - y(0)$$

and (from the integral equation): $y(o) = 1$

sam

3. Jul 17, 2010

### vela

Staff Emeritus
You took the Laplace transform of the exponentials incorrectly. You need the reciprocal of what you have. It should be, for example,

$$L[e^{3x}] = \frac{1}{s-3}$$

4. Jul 18, 2010

### MisterMan

Thanks sam and vela, completely missed that mistake.

Sam, I'm not sure how to use the differentiation approach. Is that starting from:

$$y(x) = e^{3x} + \int_0^{x}\hspace{1mm}y(t)\hspace{1mm}e^{x-t}\hspace{1mm}dt$$

And then differentiating both with respect to x? If so, how do I go about differentiating the integral?

5. Jul 18, 2010

### vela

Staff Emeritus
6. Jul 18, 2010

### Mute

A note on your notation: $(f\ast g)(x)$ is typically used to denote a convolution, which is not what you have in the first line of what I've quoted. The integral limits on a convolution run over the entire domain. You then have the property that the fourier transform of a convolution is the product of the fourier transforms of the convoluted functions.

Your second line quoted, nevertheless, is correct. I am merely pointing out that most people would assume the first line refers to a convolution, which it really isn't, so be aware of this when you use the (f*g) notation.

7. Jul 18, 2010

### vela

Staff Emeritus
Laplace transforms, unlike Fourier transforms, are traditionally one-sided, so you can take both y(t) and g(t) to vanish for t<0. With that assumption, the convolution integral reduces to the expression MisterMan wrote.