Question Number 184683 by alcohol last updated on 10/Jan/23
$${x}''\:−\:\frac{\mathrm{5}}{{t}}\:{x}'\:+\:\frac{\mathrm{8}}{{t}^{\mathrm{2}} }\:{x}\:=\:\frac{\mathrm{2}\:{ln}\:{t}}{{t}^{\mathrm{2}} } \\ $$$${solve}\:{the}\:{differential}\:{eqn} \\ $$
Answered by qaz last updated on 10/Jan/23
$${t}^{\mathrm{2}} {x}''−\mathrm{5}{tx}'+\mathrm{8}{x}=\mathrm{2}{lnt} \\ $$$${t}={e}^{{z}} , \\ $$$${x}_{{p}} =\frac{\mathrm{1}}{{D}\left({D}−\mathrm{1}\right)−\mathrm{5}{D}+\mathrm{8}}\mathrm{2}{z}=\left(\frac{\mathrm{1}}{{D}−\mathrm{4}}−\frac{\mathrm{1}}{{D}−\mathrm{2}}\right){z} \\ $$$$={e}^{\mathrm{4}{z}} \frac{\mathrm{1}}{{D}}{e}^{−\mathrm{4}{z}} {z}−{e}^{\mathrm{2}{z}} \frac{\mathrm{1}}{{D}}{e}^{−\mathrm{2}{z}} {z}={e}^{\mathrm{4}{z}} \int{e}^{−\mathrm{4}{z}} {zdz}−{e}^{\mathrm{2}{z}} \int{e}^{−\mathrm{2}{z}} {zdz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{z}+\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$\Rightarrow{x}={C}_{\mathrm{1}} {e}^{\mathrm{4}{z}} +{C}_{\mathrm{2}} {e}^{\mathrm{2}{z}} +\frac{\mathrm{1}}{\mathrm{4}}{z}+\frac{\mathrm{3}}{\mathrm{16}}={C}_{\mathrm{1}} {t}^{\mathrm{4}} +{C}_{\mathrm{2}} {t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{lnt}+\frac{\mathrm{3}}{\mathrm{16}} \\ $$