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x-5-t-x-8-t-2-x-2-ln-t-t-2-solve-the-differential-eqn-

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Question Number 184683 by alcohol last updated on 10/Jan/23
x′′ − (5/t) x′ + (8/t^2 ) x = ((2 ln t)/t^2 ) solve the differential eqn
$${x}''\:−\:\frac{\mathrm{5}}{{t}}\:{x}'\:+\:\frac{\mathrm{8}}{{t}^{\mathrm{2}} }\:{x}\:=\:\frac{\mathrm{2}\:{ln}\:{t}}{{t}^{\mathrm{2}} } \\ $$$${solve}\:{the}\:{differential}\:{eqn} \\ $$
Answered by qaz last updated on 10/Jan/23
t^2 x′′−5tx′+8x=2lnt t=e^z , x_p =(1/(D(D−1)−5D+8))2z=((1/(D−4))−(1/(D−2)))z =e^(4z) (1/D)e^(−4z) z−e^(2z) (1/D)e^(−2z) z=e^(4z) ∫e^(−4z) zdz−e^(2z) ∫e^(−2z) zdz =(1/4)z+(3/(16)) ⇒x=C_1 e^(4z) +C_2 e^(2z) +(1/4)z+(3/(16))=C_1 t^4 +C_2 t^2 +(1/4)lnt+(3/(16))
$${t}^{\mathrm{2}} {x}''−\mathrm{5}{tx}'+\mathrm{8}{x}=\mathrm{2}{lnt} \\ $$$${t}={e}^{{z}} , \\ $$$${x}_{{p}} =\frac{\mathrm{1}}{{D}\left({D}−\mathrm{1}\right)−\mathrm{5}{D}+\mathrm{8}}\mathrm{2}{z}=\left(\frac{\mathrm{1}}{{D}−\mathrm{4}}−\frac{\mathrm{1}}{{D}−\mathrm{2}}\right){z} \\ $$$$={e}^{\mathrm{4}{z}} \frac{\mathrm{1}}{{D}}{e}^{−\mathrm{4}{z}} {z}−{e}^{\mathrm{2}{z}} \frac{\mathrm{1}}{{D}}{e}^{−\mathrm{2}{z}} {z}={e}^{\mathrm{4}{z}} \int{e}^{−\mathrm{4}{z}} {zdz}−{e}^{\mathrm{2}{z}} \int{e}^{−\mathrm{2}{z}} {zdz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{z}+\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$\Rightarrow{x}={C}_{\mathrm{1}} {e}^{\mathrm{4}{z}} +{C}_{\mathrm{2}} {e}^{\mathrm{2}{z}} +\frac{\mathrm{1}}{\mathrm{4}}{z}+\frac{\mathrm{3}}{\mathrm{16}}={C}_{\mathrm{1}} {t}^{\mathrm{4}} +{C}_{\mathrm{2}} {t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{lnt}+\frac{\mathrm{3}}{\mathrm{16}} \\ $$

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