Question

$laplace\:e^{-2t}\sin ^2\left(t\right)$

how to solve it

Answer 1

$\mathrm{Laplace\:Transform\:of\:}e^{-2t}\sin ^2\left(t\right):\quad \frac{2}{\left(s+2\right)\left(s^2+4s+8\right)}$

$\mathrm{Use\:the\:following\:identity}:\quad \sin ^2\left(x\right)=\frac{1}{2}-\cos \left(2x\right)\frac{1}{2}$

$=L\left\{\left(\frac{1}{2}-\cos \left(2t\right)\frac{1}{2}\right)e^{-2t}\right\}$

$\mathrm{Expand\:}\left(\frac{1}{2}-\cos \left(2t\right)\frac{1}{2}\right)e^{-2t}:\quad \frac{1}{2}e^{-2t}-\frac{1}{2}e^{-2t}\cos \left(2t\right)$

$=L\left\{\frac{1}{2}e^{-2t}-\frac{1}{2}e^{-2t}\cos \left(2t\right)\right\}$

$\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform:}$

$\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b:\quad L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}$

$=\frac{1}{2}L\left\{e^{-2t}\right\}-\frac{1}{2}L\left\{e^{-2t}\cos \left(2t\right)\right\}$

$=\frac{1}{2}\cdot \frac{1}{s+2}-\frac{1}{2}\cdot \frac{s+2}{\left(s+2\right)^2+4}$

$\mathrm{Simplify\:}\frac{1}{2}\frac{1}{s+2}-\frac{1}{2}\frac{s+2}{\left(s+2\right)^2+4}:\quad \frac{2}{\left(s+2\right)\left(s^2+4s+8\right)}$

$=\frac{2}{\left(s+2\right)\left(s^2+4s+8\right)}$

Answer 2

By using the above Laplace transform calculator, we convert a function f(t) from the time domain, to a function F(s) of the complex variable s.

The Laplace transform provides us with a complex function of a complex variable. This may not have significant meaning to us at face value, but Laplace transforms are extremely useful in mathematics, engineering, and science