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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 $y(t)$

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  • $\begingroup$ 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. $\endgroup$ Apr 16, 2014 at 3:18
  • $\begingroup$ Thanks! I will. I just don't have any programming knowledge. Thanks again. $\endgroup$
    – Jay
    Apr 16, 2014 at 3:20
  • $\begingroup$ I can't wait to see what you come up with :o) $\endgroup$ Apr 16, 2014 at 3:21
  • $\begingroup$ 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. $\endgroup$ Apr 16, 2014 at 3:24
  • $\begingroup$ That looks much better. Thanks. But are you able to help me with the question itself?! $\endgroup$
    – Jay
    Apr 16, 2014 at 3:26

1 Answer 1

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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

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

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