Question Number 221973 by ajfour last updated on 14/Jun/25
$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{y}={k}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{6}}{{x}^{\mathrm{4}} }\:\:\:\:\:\: \\ $$$${Find}\:{y}\left({x}\right)\:\:\:\:\left({k}\:{is}\:{constant}\right). \\ $$
Answered by mahdipoor last updated on 14/Jun/25
$${get}\:\:{y}_{{p}} ={a}+\frac{{b}}{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{eq}:\:\left(\frac{\mathrm{6}{b}}{{x}^{\mathrm{4}} }\right)+\left({a}+\frac{{b}}{{x}^{\mathrm{2}} }\right)={k}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{6}}{{x}^{\mathrm{4}} } \\ $$$$\Rightarrow{a}={k}\:\:\:\:{b}=−\mathrm{1} \\ $$$${get}\:\:\:{y}={y}_{{p}} +{y}_{{q}} \\ $$$$\Rightarrow{eq}:\:\:{y}_{{q}} ^{''} +{y}_{{q}} =\mathrm{0}\:\Rightarrow\:{y}_{{q}} ={Ae}^{{Bx}} \:\Rightarrow\:{Ae}^{{Bx}} \left(\mathrm{1}+{B}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${B}={i} \\ $$$$\Rightarrow \\ $$$${y}={k}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{Ae}^{{ix}} ={k}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{A}\left({cosx}+{isinx}\right) \\ $$