Laplace transform of integral equation

Question

Use Laplace transforms to solve the integral equation

$$y(t)-\frac{1}{2}\int_0^ty(t-v)~dv=1$$

First find the Laplace transform $Y(s)$ of $y(t)$

Hi Jay. This is difficult to understand. Is the .5 supposed to be $5$ or $0.5$? Why not try formatting it yourself? Start maths-mode with a dollar sign and end maths-mode with a dollar sign. Then, inside maths-mode, you can type stuff like \int_0^{\infty} to get $\int_0^{\infty}$. If you want it big on a line of its own, start with two dollar signs and end the line with two dollar signs. — Fly by Night, Apr 16 '14 at 03:18
Thanks! I will. I just don't have any programming knowledge. Thanks again. — Jay, Apr 16 '14 at 03:20
Nice work. I've tidied up a little bit. Click edit to see what I have done and that might help you learn how to format. Just click cancel to close the edit window. — Fly by Night, Apr 16 '14 at 03:24
That looks much better. Thanks. But are you able to help me with the question itself?! — Jay, Apr 16 '14 at 03:26
I really don't know where to start, nor do I really know how to type it out to explain it. — Jay, Apr 16 '14 at 03:39

score 1 · Accepted Answer · answered Apr 16 '14 at 04:09

Taking Laplace transform of the integral equation gives

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

Now, all you need to do is to find the inverse Laplace transform which will give you the solution

$$ y(t) = e^{t/2} . $$

Note: The integral

$$ \int_{0}^{t}y(t-v)dv $$

is the convolution of the functions $1$ and $y(t)$ and we used the fact

$$ \mathcal{L}(f*g) = \mathcal{L}(f)\mathcal{L}(g). $$

1 Answers1