Question Number 1056 by 123456 last updated on 25/May/15
$${y}''+{y}=\mathrm{8}{cos}\left({t}\right) \\ $$$${y}\left(\mathrm{0}\right)=+\mathrm{1} \\ $$$${y}'\left(\mathrm{0}\right)=−\mathrm{1} \\ $$
Commented by prakash jain last updated on 26/May/15
$$\mathrm{characterstic}\:\mathrm{equation} \\ $$$${r}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${y}_{{h}} ={c}_{\mathrm{1}} \mathrm{cos}\:{t}+{c}_{\mathrm{2}} \mathrm{sin}\:{t} \\ $$
Answered by prakash jain last updated on 26/May/15
$${y}_{{p}} ={kt}\mathrm{sin}\:{t} \\ $$$${y}_{{p}} '={kt}\mathrm{cos}\:{t}+{k}\mathrm{sin}\:{t} \\ $$$${y}_{{p}} ''=−{kt}\mathrm{sin}\:{t}+{k}\mathrm{cos}\:{t}+{k}\mathrm{cos}\:{t} \\ $$$${y}_{{p}} ''+{y}=\mathrm{2}{k}\mathrm{cos}\:{t}=\mathrm{8cos}\:{t}\Rightarrow{k}=\mathrm{4} \\ $$$${y}={y}_{{p}} +{y}_{{h}} ={c}_{\mathrm{1}} \mathrm{cos}\:{t}+{c}_{\mathrm{2}} \mathrm{sin}\:{t}+\mathrm{4}{t}\mathrm{cos}\:{t} \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow{c}_{\mathrm{1}} =\mathrm{1} \\ $$$${y}'=\mathrm{4}{t}\mathrm{cos}\:{t}+\mathrm{4sin}\:{t}−{c}_{\mathrm{1}} \mathrm{sin}\:{t}+{c}_{\mathrm{2}} \mathrm{cos}\:{t} \\ $$$${y}'\left(\mathrm{0}\right)=−\mathrm{1}\Rightarrow{c}_{\mathrm{2}} =−\mathrm{1} \\ $$$${y}=\mathrm{cos}\:{t}−\mathrm{sin}\:{t}+\mathrm{4}{t}\mathrm{sin}\:{t} \\ $$