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Question-88181




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

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