Use-Laplace-transform-to-solve-the-differential-equation-d-2-v-t-dt-2-6-dv-t-dt-8v-t-2u-t-when-v-0-1-and-v-0-2-

Question Number 184936 by MikeH last updated on 14/Jan/23

$$\mathrm{Use}\:\mathrm{Laplace}\:\mathrm{transform}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation} \\ $$$$\:\frac{{d}^{\mathrm{2}} {v}\left({t}\right)}{{dt}^{\mathrm{2}} }\:+\mathrm{6}\frac{{dv}\left({t}\right)}{{dt}}\:+\:\mathrm{8}{v}\left({t}\right)\:=\:\mathrm{2}{u}\left({t}\right)\:\: \\ $$$$\mathrm{when}\:{v}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:\mathrm{and}\:\overset{\bullet} {{v}}\left(\mathrm{0}\right)\:=\:−\mathrm{2} \\ $$

Answered by hmr last updated on 15/Jan/23

L{v′′} + 6L{v′} + 8L{v} = 2L{u} → (s^2 L{v} − sv(0) −v′(0)) + 6(sL{v} − v(0)) + 8L{v} = 2L{u} → (s^2 + 6s + 8)L{v}−s+2−6 = 2L{u} → (s^2 + 6s + 8)L{v} = 2L{u} + s + 4 → L{v} = ((2L{u} + s + 4)/(s^2 + 6s + 8)) → L{v} = ((2L{u} + (s + 4))/((s + 4)(s + 2))) → L{v} = ((2L{u})/((s + 4)(s + 2))) + (1/(s + 2)) → L{v} = ((2L{u})/((s + 4)(s + 2))) + L{e^(−2t) } → L{v} = ((4L{u})/(s + 4)) − ((2L{u})/(s + 2)) + L{e^(−2t) } → L{v} = 4L{e^(−4t) } L{u} − 2L{e^(−2t) } L{u}+ L{e^(−2t) } Sorry that′s all I know.

$${L}\left\{{v}''\right\}\:+\:\mathrm{6}{L}\left\{{v}'\right\}\:+\:\mathrm{8}{L}\left\{{v}\right\}\:=\:\mathrm{2}{L}\left\{{u}\right\} \\ $$$$\rightarrow\: \\ $$$$\left({s}^{\mathrm{2}} \:{L}\left\{{v}\right\}\:−\:{sv}\left(\mathrm{0}\right)\:−{v}'\left(\mathrm{0}\right)\right)\: \\ $$$$+\:\mathrm{6}\left({sL}\left\{{v}\right\}\:−\:{v}\left(\mathrm{0}\right)\right) \\ $$$$+\:\mathrm{8}{L}\left\{{v}\right\}\:=\:\mathrm{2}{L}\left\{{u}\right\} \\ $$$$\rightarrow \\ $$$$\left({s}^{\mathrm{2}} \:+\:\mathrm{6}{s}\:+\:\mathrm{8}\right){L}\left\{{v}\right\}−{s}+\mathrm{2}−\mathrm{6}\:=\:\mathrm{2}{L}\left\{{u}\right\} \\ $$$$\rightarrow \\ $$$$\left({s}^{\mathrm{2}} \:+\:\mathrm{6}{s}\:+\:\mathrm{8}\right){L}\left\{{v}\right\}\:=\:\mathrm{2}{L}\left\{{u}\right\}\:+\:{s}\:+\:\mathrm{4} \\ $$$$\rightarrow \\ $$$${L}\left\{{v}\right\}\:=\:\frac{\mathrm{2}{L}\left\{{u}\right\}\:+\:{s}\:+\:\mathrm{4}}{{s}^{\mathrm{2}} \:+\:\mathrm{6}{s}\:+\:\mathrm{8}} \\ $$$$\rightarrow \\ $$$${L}\left\{{v}\right\}\:=\:\frac{\mathrm{2}{L}\left\{{u}\right\}\:+\:\left({s}\:+\:\mathrm{4}\right)}{\left({s}\:+\:\mathrm{4}\right)\left({s}\:+\:\mathrm{2}\right)} \\ $$$$\rightarrow \\ $$$${L}\left\{{v}\right\}\:=\:\:\frac{\mathrm{2}{L}\left\{{u}\right\}}{\left({s}\:+\:\mathrm{4}\right)\left({s}\:+\:\mathrm{2}\right)}\:+\:\frac{\mathrm{1}}{{s}\:+\:\mathrm{2}} \\ $$$$\rightarrow \\ $$$${L}\left\{{v}\right\}\:=\:\:\frac{\mathrm{2}{L}\left\{{u}\right\}}{\left({s}\:+\:\mathrm{4}\right)\left({s}\:+\:\mathrm{2}\right)}\:+\:{L}\left\{{e}^{−\mathrm{2}{t}} \right\} \\ $$$$\rightarrow \\ $$$${L}\left\{{v}\right\}\:=\:\:\frac{\mathrm{4}{L}\left\{{u}\right\}}{{s}\:+\:\mathrm{4}}\:−\:\frac{\mathrm{2}{L}\left\{{u}\right\}}{{s}\:+\:\mathrm{2}}\:+\:{L}\left\{{e}^{−\mathrm{2}{t}} \right\} \\ $$$$\rightarrow \\ $$$${L}\left\{{v}\right\}\:=\:\:\mathrm{4}{L}\left\{{e}^{−\mathrm{4}{t}} \right\}\:{L}\left\{{u}\right\}\:−\:\mathrm{2}{L}\left\{{e}^{−\mathrm{2}{t}} \right\}\:{L}\left\{{u}\right\}+\:{L}\left\{{e}^{−\mathrm{2}{t}} \right\} \\ $$$${Sorry}\:{that}'{s}\:{all}\:{I}\:{know}. \\ $$

Commented by MikeH last updated on 15/Jan/23