MHB Laplace Convolution: f(t)=-5t^2+9

Alex2124
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f(t)=-5t^2+9\int_{0}^{t} \,f(t-u)sin(9u)du
 

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Alex2124 said:
f(t)=-5t^2+9\int_{0}^{t} \,f(t-u)sin(9u)du

$\displaystyle \mathcal{L} \left\{ f\left( t \right) \right\} = F\left( s \right) $, so

$\displaystyle \begin{align*} \mathcal{L} \left\{ f\left( t \right) \right\} &= \mathcal{L}\left\{ -5\,t^2 \right\} + 9\,\mathcal{L}\left\{ \int_0^t{ f\left( t - u \right) \,\sin{\left( 9\,u \right) } \,\mathrm{d}u } \right\} \\
F\left( s \right) &= -5 \left( \frac{2}{s^3} \right) + 9 \,F\left( s \right) \left( \frac{9}{s^2 + 81} \right) \end{align*}$

Now solve for $F\left( s \right) $.
 


I find this equation to be quite interesting. It looks like a combination of a quadratic function and an integral. I'm curious to know what the function f(t) represents and how it relates to the integral in the equation. Can you provide any more context or information about this equation?
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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