Question Number 88181 by ubaydulla last updated on 08/Apr/20
Commented by mathmax by abdo last updated on 09/Apr/20
$$\left.\mathrm{5}\right)\:{y}^{'} \:+{y}\:={e}^{\mathrm{2}{x}} \:\:\:\left({he}\right)\rightarrow{y}^{'} \:+{y}\:=\mathrm{0}\Rightarrow\frac{{y}^{'} }{{y}}=−\mathrm{1}\:\Rightarrow{ln}\mid{y}\mid=−{x}\:+{c}\:\Rightarrow \\ $$$${y}\:={k}\:{e}^{−{x}} \:\:\:\:\:{mvc}\:{method}\:\:{y}^{'} \:={k}^{'} \:{e}^{−{x}} \:−{ke}^{−{x}} \\ $$$$\left({e}\right)\Rightarrow{k}^{'} \:{e}^{−{x}} −{ke}^{−{x}} +{ke}^{−{x}} \:={e}^{\mathrm{2}{x}} \:\Rightarrow{k}^{'} \:={e}^{\mathrm{3}{x}} \:\Rightarrow{k}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{x}} \:+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\left(\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{x}} \:+\lambda\right){e}^{−{x}} \:=\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{2}{x}} \:+\lambda{e}^{−{x}} \\ $$