Question

Determine L^-1{e^s(s²-1/(s²+1)²)}

Answer 1

$L^{-1}\left\{ e^s\left( \frac{s^2-1}{\left( s^2+1 \right) ^2} \right) \right\} =L^{-1}\left( \frac{e^s}{s^2+1}-2\frac{e^s}{\left( s^2+1 \right) ^2} \right) \\L^{-1}\left( \frac{1}{s^2+1} \right) =\sin t\Rightarrow L^{-1}\left( \frac{e^s}{s^2+1} \right) =\sin \left( t+1 \right) \\L^{-1}\left( -\frac{2s}{\left( s^2+1 \right) ^2} \right) =L^{-1}\left( \left( \frac{1}{s^2+1} \right) ' \right) =t\sin t\\L^{-1}\left( \frac{1}{\left( s^2+1 \right) ^2} \right) =-\frac{1}{2}L^{-1}\left( \frac{\frac{2s}{\left( s^2+1 \right)}}{s} \right) =-\frac{1}{2}\int_0^t{\left( -x\sin x \right) dx}=\\=-\frac{1}{2}\left( x\cos x|_{0}^{t}-\int_0^t{\cos xdx} \right) =\frac{1}{2}\left( -t\cos t+\sin t \right) \\L^{-1}\left( \frac{e^s}{\left( s^2+1 \right) ^2} \right) =\frac{1}{2}\left( -\left( t+1 \right) \cos \left( t+1 \right) +\sin \left( t+1 \right) \right) \\L^{-1}\left\{ e^s\left( \frac{s^2-1}{\left( s^2+1 \right) ^2} \right) \right\} =\sin \left( t+1 \right) -2\cdot \frac{1}{2}\left( -\left( t+1 \right) \cos \left( t+1 \right) +\sin \left( t+1 \right) \right) =\\=\left( t+1 \right) \cos \left( t+1 \right)$