Question Number 149274 by BHOOPENDRA last updated on 04/Aug/21
Answered by Olaf_Thorendsen last updated on 05/Aug/21
$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{3}\frac{{dy}}{{dx}}+\mathrm{2}{y}\:=\:{e}^{{x}} \:\:\:\:\left(\mathrm{E}\right) \\ $$$$\mathrm{Let}\:{y}\:=\:{e}^{{x}} {u},\:{y}'\:=\:{e}^{{x}} \left({u}'+{u}\right),\:{y}''\:=\:{e}^{{x}} \left({u}''+\mathrm{2}{u}'+{u}\right) \\ $$$$\left(\mathrm{E}\right)\::\:{u}''+\mathrm{2}{u}'+{u}−\mathrm{3}\left({u}'+{u}\right)+\mathrm{2}{u}\:=\:\mathrm{1} \\ $$$$\left(\mathrm{E}\right)\::\:{u}''−{u}'\:=\:\mathrm{1} \\ $$$$\Rightarrow\:{u}'\:=\:−\mathrm{1},\:{u}\:=\:{a}−{x} \\ $$$${y}\:=\:{e}^{{x}} {u}\:=\:\left({a}−{x}\right){e}^{{x}} \\ $$