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Here is the problem

Solve $y'(t) = 1 - \int_{0}^{t} y(t - v)e^{-2v}dv$

The solution sets $\mathcal{L}(y) = Y(s)$ and does the following

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Notice that in step 1, they have $Y(s)\dfrac{1}{s+2}$

Are they implying $\mathcal{L}(y(t) * e^{-2t}) = \mathcal{L}(y(t)) \mathcal{L}(e^{-2t}) = Y(s)\mathcal{L}(e^{-2t}) = Y(s)\dfrac{1}{s+2}$

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    fixed mistake on exp(-2t). Thank you for catching that svenkatr2012-06-14

1 Answers 1

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Yes they are implying that. Applying the Laplace transform to a convolution gives a product. Write

$\mathcal{L}\{f*g\}(s):=\int_0^\infty \left(\int_0^x f(u)g(x-u)du\right)e^{-sx}dx=\iint_D f(u)g(x-u)e^{-sx}dudx$

The region of integration in the $xu$-plane is $D=\{(x,u):0\le u\le x\}$, an infinite triangle. Writing out the substitution $v=x-u$ our region of integration in the $uv$-plane is simply $(0,\infty)\times(0,\infty)$, so

$=\iint_{(0,\infty)^2} f(u)g(v)e^{-s(u+v)}dudv=\int_0^\infty f(u)e^{-su}du\int_0^\infty g(v)e^{-sv}dv=\mathcal{L}\{f\}(s)\cdot \mathcal{L}\{g\}(s).$

Note the Jacobian determinant of the transformation $(x,u)\mapsto(u,x-u)$ is simply $1$ (in abs. value).

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    @jak: What cases are you talking about? The triangular region comes from the definition of convolution and the Laplace transform being put together, and the infinite square comes from the change of variables: none of that has anything to do with choice of $f$ and $g$ (except as far as regularity and therefore convergence is concerned).2012-06-14