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Question Number 170329 by ali009 last updated on 21/May/22
$${find}\:{the}\:{general}\:{solution}\:{of}\: \\ $$$$\left.\mathrm{1}\right){y}'''−\mathrm{7}{y}'+\mathrm{6}{y}={x} \\ $$$$\left.\mathrm{2}\right){y}'''−\mathrm{3}{y}''+\mathrm{2}{y}'=\frac{{e}^{{x}} }{\mathrm{1}+{e}^{−{x}} } \\ $$
Answered by LEKOUMA last updated on 22/May/22
$$\left.\mathrm{1}\right)\:{y}^{'''} −\mathrm{7}{y}'+\mathrm{6}{y}={x} \\ $$$${Eh} \\ $$$${y}'''−\mathrm{7}{y}'+\mathrm{6}{y}=\mathrm{0} \\ $$$${Ec} \\ $$$${r}^{\mathrm{3}} −\mathrm{7}{r}+\mathrm{6}=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left({r}^{\mathrm{2}} +{r}−\mathrm{6}\right)=\mathrm{0} \\ $$$${r}−\mathrm{1}=\mathrm{0}\:{or}\:{r}^{\mathrm{2}} +{r}−\mathrm{6}=\mathrm{0} \\ $$$$\:{r}_{\mathrm{1}} =\mathrm{1}\:{or}\:\:{r}_{\mathrm{2}} =−\mathrm{3}\:,\:{r}_{\mathrm{3}} =\mathrm{2} \\ $$$${y}={c}_{\mathrm{1}} {e}^{{x}} +\:{c}_{\mathrm{2}} {e}^{−\mathrm{3}{x}} +{c}_{\mathrm{3}} {e}^{\mathrm{2}{x}} \\ $$$${Sp} \\ $$$${y}_{{p}} ={x}\:\Rightarrow\:{y}_{{p}} ={ax} \\ $$$${y}_{{p}} '={a} \\ $$$${y}_{{p}} ''=\mathrm{0} \\ $$$${y}_{{p}} '''=\mathrm{0} \\ $$$$\mathrm{0}−\mathrm{7}{a}+\mathrm{6}{ax}={x} \\ $$$$\mathrm{6}{a}=\mathrm{1}\:\Leftrightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{\mathrm{6}}{x} \\ $$$${y}\left({x}\right)={y}_{{h}} +{y}_{{p}} \\ $$$${y}\left({x}\right)={c}_{\mathrm{1}} {e}^{{x}} +{c}_{\mathrm{2}} {e}^{−\mathrm{3}{x}} +{c}_{\mathrm{3}} {e}^{\mathrm{2}{x}} +\frac{\mathrm{1}}{\mathrm{6}}{x} \\ $$$${c}_{\mathrm{1}} ,{c}_{\mathrm{2}} \:{et}\:{c}_{\mathrm{3}} \:{cste} \\ $$