Question Number 129268 by BHOOPENDRA last updated on 14/Jan/21
$$\frac{\mathrm{1}}{\left({s}^{\mathrm{2}} +\mathrm{2}{s}+\mathrm{5}\right)^{\mathrm{2}} }\:{find}\:{inverse}\:{laplace} \\ $$
Answered by mnjuly1970 last updated on 14/Jan/21
![F(s)=(1/(((s+1)^2 +2^2 )^2 )) we know: L^( −1) ((1/((s+1)^2 +2^2 )))=(1/2)e^(−t) sin(2t) ∴ L^(−1) [((1/((s+1)^2 +2^2 )) )^2 ]=(1/4)∫_0 ^( t) (e^(−λ) sin(2λ)).(e^(−(t−λ)) sin(2(t−λ))dλ =(1/4)∫_0 ^( t) e^(−t) sin(2λ).sin(2t−2λ)dλ =(1/8)∫_0 ^( t) e^(−t) [cos(2t−2λ−2λ)−cos(2t−2λ+2λ)]dλ =(1/8){∫_0 ^( t) e^(−t) cos(4λ−2t)dλ−∫_0 ^( t) e^(−t) cos(2t)dλ =...](https://www.tinkutara.com/question/Q129280.png)
$$\:\mathscr{F}\left({s}\right)=\frac{\mathrm{1}}{\left(\left({s}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:{we}\:{know}:\:\mathscr{L}^{\:\:−\mathrm{1}} \left(\frac{\mathrm{1}}{\left({s}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{t}} {sin}\left(\mathrm{2}{t}\right) \\ $$$$\therefore\:\mathscr{L}\:^{−\mathrm{1}} \left[\left(\frac{\mathrm{1}}{\left({s}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }\:\right)^{\mathrm{2}} \:\right]=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:{t}} \left({e}^{−\lambda} {sin}\left(\mathrm{2}\lambda\right)\right).\left({e}^{−\left({t}−\lambda\right)} {sin}\left(\mathrm{2}\left({t}−\lambda\right)\right){d}\lambda\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:{t}} {e}^{−{t}} {sin}\left(\mathrm{2}\lambda\right).{sin}\left(\mathrm{2}{t}−\mathrm{2}\lambda\right){d}\lambda \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:{t}} {e}^{−{t}} \left[{cos}\left(\mathrm{2}{t}−\mathrm{2}\lambda−\mathrm{2}\lambda\right)−{cos}\left(\mathrm{2}{t}−\mathrm{2}\lambda+\mathrm{2}\lambda\right)\right]{d}\lambda \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left\{\int_{\mathrm{0}} ^{\:{t}} {e}^{−{t}} {cos}\left(\mathrm{4}\lambda−\mathrm{2}{t}\right){d}\lambda−\int_{\mathrm{0}} ^{\:{t}} {e}^{−{t}} {cos}\left(\mathrm{2}{t}\right){d}\lambda\right. \\ $$$$=… \\ $$$$ \\ $$
Commented by BHOOPENDRA last updated on 14/Jan/21
$${what}\:\lambda\:{represent}\:{here}? \\ $$