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I'm working through some lectures notes on Differential Equations, and in the Laplace Transform section I've encountered the following problem:

Problem

Find a solution to

$$x'(t)+\int^{t}_{0}(t-s)x(s)ds=t+\frac{1}{2}t^2+\frac{1}{24}t^4$$

I'm not really sure how we can apply the Laplace transform here; any assisstance would be greatly appreciated. Regards as always, MM.

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The Laplace transform of a convolution $f\star g$ is the product of the Laplace transforms of $f$ and $g$. So ${\cal L} \int_0^t g(t-s) f(s) ds=F(s)G(s)$. –  David Mitra Jan 12 '12 at 13:58
    
@DavidMitra: Here then, are we to assume that $g$ is the identity map (as $g(t-s)=t-s$)? –  Mathmo Jan 12 '12 at 14:16

1 Answer 1

up vote 4 down vote accepted

Taking the Laplace transform of the left hand side, using the convolution formula on the integral (note the integral is $f\star g$ where $f $ is the identity function and and $g=x$): $${\cal L} \Bigl( x'(t)+\int^{t}_{0}(t-s)x(s)ds\Bigr) = s X(s)-x(0) + {1\over s^2}\cdot X(s). $$

Taking the Laplace transform of the right hand side: $$ {\cal L}\Bigl( t+\frac{1}{2}t^2+\frac{1}{24}t^4 \Bigr)= {1\over s^2}+{1\over 2}{2\over s^3}+{1\over 24}\cdot{24\over s^5}. $$ So, we have: $$ s X(s)-x(0) + {1\over s^2}\cdot X(s)= {1\over s^2}+ {1\over s^3}+ {1\over s^5}. $$

Now solve for $X(s)$ and then take the inverse transform to find $x(t)$.

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