I keep seeing videos about the laplace of transform of the property itself but not how to "apply" the property, specially in a case where the problem does not look anything like the property. In the problem given there is no "e" in-fact there is a sin instead , my math is beyond weak , so i do not see the relation or anything this entire subject is foreign language to me. If someone can help paint a clearer picture.
I know I have a clear lack of knowledge in this because anything something does not match a given example or exactly how the book has it then I am at a loss of how any minor change affects the entire problem.
One issue could be that I think some of this stuff requires a background in differential equations that I do not have. Not sure why I was allowed to take this course but here I am.
See the second line of the display here: $$ \sin(z) = \frac{\mathrm{e}^{\mathrm{i}z} - \mathrm{e}^{-\mathrm{i}z}}{2\mathrm{i}} \mathrm{.} $$
So using the above and linearity, \begin{align*} \mathcal{L}(10t \sin(100t)) &= \mathcal{L} \left( 10 t \frac{\mathrm{e}^{\mathrm{i}(100t)} - \mathrm{e}^{-\mathrm{i}(100t)}}{2\mathrm{i}} \right) \\ &= \frac{10}{2\mathrm{i}} \mathcal{L} \left( t \left( \mathrm{e}^{\mathrm{i}(100t)} - \mathrm{e}^{-\mathrm{i}(100t)} \right) \right) \\ &= \frac{5}{\mathrm{i}} \mathcal{L} \left( t \mathrm{e}^{(100\mathrm{i})t} \right) - \frac{5}{\mathrm{i}} \mathcal{L} \left(t \mathrm{e}^{(-100\mathrm{i})t} \right) \text{.} \end{align*}