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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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    @DavidMitra: Here then, are we to assume that $g$ is the identity map (as $g(t-s)=t-s$)?2012-01-12

1 Answers 1

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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 $x(t)$.