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Question Number 112840 by bemath last updated on 10/Sep/20
$$\mathrm{solve}\:\mathrm{y}''−\mathrm{y}'+\mathrm{e}^{\mathrm{2x}} \mathrm{y}\:=\:\mathrm{0} \\ $$
Answered by john santu last updated on 10/Sep/20
$$\:{solve}\:{y}''−{y}'\:+{e}^{\mathrm{2}{x}} \:{y}\:=\:\mathrm{0}. \\ $$$$\:{substitute}\:{u}\:=\:{e}^{{x}} \\ $$$$\begin{cases}{\frac{{dy}}{{dx}}\:=\:{e}^{{x}} \:=\:{u}}\\{\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\:\frac{{d}}{{dx}}\left({u}\:\frac{{dy}}{{du}}\right)=\:{u}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{du}^{\mathrm{2}} }\:+{u}\:\frac{{dy}}{{du}}}\end{cases} \\ $$$${Hence}\:{y}''−{y}'\:+\:{e}^{\mathrm{2}{x}} {y}\:=\: \\ $$$$\left({u}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{du}^{\mathrm{2}} }\:+\:{u}\:\frac{{dy}}{{dx}}\right)−\left({u}\:\frac{{dy}}{{du}}\right)+{u}^{\mathrm{2}} {y}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{u}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{du}^{\mathrm{2}} }\:+{u}^{\mathrm{2}} {y}\:=\:\mathrm{0} \\ $$$${so}\::\:{u}^{\mathrm{2}} \left(\frac{{d}^{\mathrm{2}} {y}}{{du}^{\mathrm{2}} }\:+\:{y}\right)\:=\:\mathrm{0} \\ $$$${because}\:{u}\:=\:{e}^{{x}} \:,\:{u}\:\neq\:\mathrm{0}\:,\:{so}\:{we}\:{can}\: \\ $$$${divide}\:{by}\:{u}^{\mathrm{2}} \:\Rightarrow\:\frac{{d}^{\mathrm{2}} {y}}{{du}^{\mathrm{2}} }\:+\:{y}\:=\:\mathrm{0} \\ $$$${homogenous}\:{solution}\: \\ $$$$\lambda^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0}\:;\:\lambda\:=\:\pm\:{i}\: \\ $$$${y}\:=\:{C}_{\mathrm{1}} \:\mathrm{cos}\:{u}\:+\:{C}_{\mathrm{2}} \mathrm{sin}\:{u} \\ $$$${y}\:=\:{C}_{\mathrm{1}} \mathrm{cos}\:\left({e}^{{x}} \right)\:+\:{C}_{\mathrm{2}} \mathrm{sin}\:\left({e}^{{x}} \right) \\ $$$$\:\:\:\:\:\frac{{JS}}{{a}\:{math}\:{farmer}} \\ $$