Question Number 16475 by Sai dadon. last updated on 22/Jun/17
$${use}\:{L}\left\{{f}'\left({t}\right)\right\}\:{to}\:{find}\:{laplace}\: \\ $$$${transform}\:{of}\:{d}^{\mathrm{2}} {x}/{dt}^{\mathrm{2}} +{n}^{\mathrm{2}} {x}={kcoswt} \\ $$$${given}\:{that}\:{x}=\mathrm{0}\:{and}\:{dx}/{dt}=\mathrm{0}\:{when}\:{t}=\mathrm{0} \\ $$$${PLZ}\:{HELP}. \\ $$$$ \\ $$
Answered by sma3l2996 last updated on 23/Jun/17
$${we}\:{know}\:{that}\:{L}\left\{{f}'\left({t}\right)\right\}={sL}\left\{{f}\left({t}\right)\right\}−{f}\left(\mathrm{0}\right)\:\:{and}\:\:{L}\left\{{f}''\left({t}\right)\right\}={sL}\left\{{f}'\left({t}\right)\right\}−{f}'\left(\mathrm{0}\right) \\ $$$${so}\:\:{L}\left\{{f}''\left({t}\right)\right\}={s}^{\mathrm{2}} {L}\left\{{f}\left({t}\right)\right\}−{sf}\left(\mathrm{0}\right)−{f}'\left(\mathrm{0}\right) \\ $$$${let}\:{x}\left({t}\right)={f}\left({t}\right) \\ $$$${L}\left\{\frac{{d}^{\mathrm{2}} {x}\left({t}\right)}{{dt}^{\mathrm{2}} }\right\}={s}^{\mathrm{2}} {L}\left\{{x}\left({t}\right)\right\}−{sx}\left(\mathrm{0}\right)−\frac{{dx}\left(\mathrm{0}\right)}{{dt}}={s}^{\mathrm{2}} {L}\left\{{x}\left({t}\right)\right\} \\ $$$${and}\:{also}\:{we}\:{have}: \\ $$$${L}\left\{\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+{n}^{\mathrm{2}} {x}\right\}={L}\left\{{kcos}\left({wt}\right)\right\} \\ $$$$\Leftrightarrow{L}\left\{\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\right\}+{n}^{\mathrm{2}} {L}\left\{{x}\right\}={kL}\left\{{cos}\left({wt}\right)\right\}={k}.\frac{{s}}{{s}^{\mathrm{2}} +{w}^{\mathrm{2}} } \\ $$$${s}^{\mathrm{2}} {L}\left\{{x}\right\}+{n}^{\mathrm{2}} {L}\left\{{x}\right\}={k}.\frac{{s}}{{s}^{\mathrm{2}} +{w}^{\mathrm{2}} } \\ $$$$\left({s}^{\mathrm{2}} +{n}^{\mathrm{2}} \right).{L}\left\{{x}\right\}={k}.\frac{{s}}{{s}^{\mathrm{2}} +{w}^{\mathrm{2}} } \\ $$$${L}\left\{{x}\right\}={k}.\frac{{s}}{\left({s}^{\mathrm{2}} +{w}^{\mathrm{2}} \right)\left({s}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)} \\ $$$$\frac{{s}}{\left({s}^{\mathrm{2}} +{w}^{\mathrm{2}} \right)\left({s}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)}=\frac{{as}+{b}}{{s}^{\mathrm{2}} +{w}^{\mathrm{2}} }+\frac{{cs}+{d}}{{s}^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$$${a}=−{c}\:,\:\frac{{b}}{{w}^{\mathrm{2}} }=−\frac{{d}}{{n}^{\mathrm{2}} }\Leftrightarrow{d}=−\left(\frac{{n}}{{w}}\right)^{\mathrm{2}} {b} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+{w}^{\mathrm{2}} \right)\left(\mathrm{1}+{n}^{\mathrm{2}} \right)}=\frac{{a}+{b}}{\mathrm{1}+{w}^{\mathrm{2}} }−\frac{{a}+\left(\frac{{n}}{{w}}\right)^{\mathrm{2}} {b}}{\mathrm{1}+{n}^{\mathrm{2}} }\Leftrightarrow\left({a}+{b}\right)\left(\mathrm{1}+{n}^{\mathrm{2}} \right)−\left({a}+\left(\frac{{n}}{{w}}\right)^{\mathrm{2}} {b}\right)\left(\mathrm{1}+{w}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$${a}\left({n}^{\mathrm{2}} −{w}^{\mathrm{2}} \right)+{b}\left(\mathrm{1}+{n}^{\mathrm{2}} −\left(\frac{{n}}{{w}}\right)^{\mathrm{2}} \left(\mathrm{1}+{w}^{\mathrm{2}} \right)\right)=\mathrm{1} \\ $$$${a}\left({n}^{\mathrm{2}} −{w}^{\mathrm{2}} \right)=\mathrm{1}+{b}\left(\frac{{n}^{\mathrm{2}} −{w}^{\mathrm{2}} }{{w}^{\mathrm{2}} }\right) \\ $$$$\left({i}\right):{a}=\frac{{b}}{{w}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{w}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}}{\left({w}^{\mathrm{2}} +\mathrm{4}\right)\left({n}^{\mathrm{2}} +\mathrm{4}\right)}=\frac{\mathrm{2}{a}+{b}}{{w}^{\mathrm{2}} +\mathrm{4}}−\frac{\mathrm{2}{a}+\left(\frac{{n}}{{w}}\right)^{\mathrm{2}} {b}}{{n}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\left(\mathrm{2}{a}+{b}\right)\left({n}^{\mathrm{2}} +\mathrm{4}\right)−\left(\mathrm{2}{a}+\left(\frac{{n}}{{w}}\right)^{\mathrm{2}} {b}\right)\left({w}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{2} \\ $$$$\mathrm{2}{a}\left({n}^{\mathrm{2}} −{w}^{\mathrm{2}} \right)+{b}\left(\mathrm{4}−\mathrm{4}\left(\frac{{n}}{{w}}\right)^{\mathrm{2}} \right)=\mathrm{2} \\ $$$$\mathrm{4}{b}\left(\frac{{w}^{\mathrm{2}} −{n}^{\mathrm{2}} }{{w}^{\mathrm{2}} }\right)=\mathrm{2}+\mathrm{2}{a}\left({w}^{\mathrm{2}} −{n}^{\mathrm{2}} \right) \\ $$$$\left({ii}\right):\frac{{b}}{{w}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}\left({w}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)}+\frac{{a}}{\mathrm{2}} \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}{a}+\frac{\mathrm{1}}{\mathrm{2}\left({w}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{w}^{\mathrm{2}} } \\ $$$${a}=\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{w}^{\mathrm{2}} }=−{c} \\ $$$$\frac{{b}}{{w}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}\left({w}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)}+\frac{\mathrm{1}}{\mathrm{2}\left({n}^{\mathrm{2}} −{w}^{\mathrm{2}} \right)}=\mathrm{0} \\ $$$${a}=−{c}=\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{w}^{\mathrm{2}} }\:{and}\:\:\:{b}={d}=\mathrm{0} \\ $$$${so}\::\:\:\:{L}\left\{{x}\right\}=\frac{{k}}{{n}^{\mathrm{2}} −{w}^{\mathrm{2}} }\left(\frac{{s}}{{s}^{\mathrm{2}} +{w}^{\mathrm{2}} }−\frac{{s}}{{s}^{\mathrm{2}} +{n}^{\mathrm{2}} }\right)=\frac{{k}}{{n}^{\mathrm{2}} −{w}^{\mathrm{2}} }\left({L}\left\{{cos}\left({wt}\right)\right\}−{L}\left\{{cos}\left({nt}\right)\right\}\right) \\ $$$${L}\left\{{x}\left({t}\right)\right\}={L}\left\{\frac{{k}}{{n}^{\mathrm{2}} −{w}^{\mathrm{2}} }\left({cos}\left({wt}\right)−{cos}\left({nt}\right)\right)\right\} \\ $$$${so}\::\:{x}\left({t}\right)=\frac{{k}\left({cos}\left({wt}\right)−{cos}\left({nt}\right)\right)}{{n}^{\mathrm{2}} −{w}^{\mathrm{2}} } \\ $$