\(\displaystyle\sqrt{{{\frac{{{t}}}{{\pi}}}}}{\cos{{2}}}{t}={t}{\frac{{{\cos{{2}}}{t}}}{{\sqrt{{\pi{t}}}}}}\)

Daniel Cormack

Answered 2021-12-13
Author has **1076** answers

\(\displaystyle{L}{\left[\sqrt{{{\frac{{{t}}}{{\pi}}}}}{\cos{{\left({2}{t}\right)}}}\right]}\)

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

\(\displaystyle-{\left({1}\right)}^{{n}}{\frac{{{d}}}{{{d}{s}}}}{\left({\frac{{{e}^{{-{\frac{{{2}}}{{{s}}}}}}}}{{\sqrt{{s}}}}}\right)}\)

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

\(\displaystyle-{\left({1}\right)}^{{n}}{\frac{{{d}}}{{{d}{s}}}}{\left({\frac{{{e}^{{-{\frac{{{2}}}{{{s}}}}}}}}{{\sqrt{{s}}}}}\right)}\)

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\(\displaystyle{L}{\left[{\left({t}^{{2}}+{4}\right)}{e}^{{{2}{t}}}-{e}^{{-{2}{t}}}{\cos{{t}}}\right]}\)