 I'm having trouble with the laplace transform: L\{\sqrt{\frac{t}{\pi}}\cos(2t)\} The problem gives Stacie Worsley 2021-12-11 Answered
I'm having trouble with the laplace transform: $$\displaystyle{L}{\left\lbrace\sqrt{{{\frac{{{t}}}{{\pi}}}}}{\cos{{\left({2}{t}\right)}}}\right\rbrace}$$
The problem gives me the transform identity $$\displaystyle{L}{\left\lbrace{\frac{{{\cos{{\left({2}{t}\right)}}}}}{{\sqrt{{\pi{t}}}}}}\right\rbrace}={\frac{{{e}^{{-{\frac{{{2}}}{{{s}}}}}}}}{{\sqrt{{s}}}}}$$ but i'm not sure/confused as to why that would help me

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Assuming t>0 (which is a usual assumption with Laplace transforms),
$$\displaystyle\sqrt{{{\frac{{{t}}}{{\pi}}}}}{\cos{{2}}}{t}={t}{\frac{{{\cos{{2}}}{t}}}{{\sqrt{{\pi{t}}}}}}$$
Not exactly what you’re looking for? Daniel Cormack
$$\displaystyle{L}{\left[\sqrt{{{\frac{{{t}}}{{\pi}}}}}{\cos{{\left({2}{t}\right)}}}\right]}$$
If you define the function: $$\displaystyle{f{{\left({t}\right)}}}=\sqrt{{{\frac{{{t}}}{{\pi}}}}}{\cos{{\left({2}{t}\right)}}}$$
If you multiply it by ($$\displaystyle{\frac{{{t}}}{{{t}}}}$$) You can rewrite: $$\displaystyle{f{{\left({t}\right)}}}={t}{\frac{{{1}}}{{\sqrt{{{t}\pi}}}}}{\cos{{\left({2}{t}\right)}}}={t}{\frac{{{\cos{{\left({2}{t}\right)}}}}}{{\sqrt{{{t}\pi}}}}}$$
At this point, I have to remind you: $$\displaystyle\text{L[t^n f(t)]=-(1)^n \frac{d}{ds^n}(L[f(t)])}$$
Finally, if you have $$\displaystyle{L}{\left[{\frac{{{\cos{{\left({2}{t}\right)}}}}}{{\sqrt{{{t}\pi}}}}}\right]}={\frac{{{e}^{{-{\frac{{{2}}}{{{s}}}}}}}}{{\sqrt{{s}}}}}$$ you have to calculate
$$\displaystyle-{\left({1}\right)}^{{n}}{\frac{{{d}}}{{{d}{s}}}}{\left({\frac{{{e}^{{-{\frac{{{2}}}{{{s}}}}}}}}{{\sqrt{{s}}}}}\right)}$$