Question

If F(s)=s^{2}+2/s(2s^{2}-7s+5) ther limt→0f(t)

Answer 1

$L^{-1}\left\{\frac{2}{s\left(2s^2-7s+5\right)}\right\}\\ =L^{-1}\left\{\frac{2}{5s}-\frac{2}{3\left(s-1\right)}+\frac{8}{15\left(2s-5\right)}\right\}\\ \frac{2}{5s}-\frac{2}{3\left(s-1\right)}+\frac{8}{15\left(2s-5\right)}\\ =L^{-1}\left\{\frac{2}{5s}-\frac{2}{3\left(s-1\right)}+\frac{8}{15\left(2s-5\right)}\right\}\\ \mathrm{Use\:the\:linearity\:property\:of\:Inverse\:Laplace\:Transform:}\\ \mathrm{For\:functions\:}f\left(s\right),\:g\left(s\right)\mathrm{\:and\:constants\:}a,\:b:\quad L^{-1}\left\{a\cdot f\left(s\right)+b\cdot g\left(s\right)\right\}=a\cdot L^{-1}\left\{f\left(s\right)\right\}+b\cdot L^{-1}\left\{g\left(s\right)\right\}\\ =L^{-1}\left\{\frac{2}{5s}\right\}-L^{-1}\left\{\frac{2}{3\left(s-1\right)}\right\}+L^{-1}\left\{\frac{8}{15\left(2s-5\right)}\right\}\\ =\frac{2}{5}\text{H}\left(t\right)\\ =\frac{2}{3}e^t\\ =\frac{4}{15}e^{\frac{5t}{2}}\\ \therefore =\frac{2}{5}\text{H}\left(t\right)-\frac{2}{3}e^t+\frac{4}{15}e^{\frac{5t}{2}}$