Laplace Convolution: f(t)=-5t^2+9

In summary, the equation given is a Laplace transform with the unknown function represented as F(s). The equation includes a quadratic function and an integral, and it is interesting to see how they are combined. The function f(t) and its relationship to the integral are unknown and further context is needed.
  • #1
Alex2124
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f(t)=-5t^2+9\int_{0}^{t} \,f(t-u)sin(9u)du
 

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  • #2
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) $.
 
  • #3


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?
 

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