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Question Number 35618 by abdo mathsup 649 cc last updated on 21/May/18
integrate the e.d. y′ +e^(−2x) y = (2x+1)cosx
$${integrate}\:{the}\:{e}.{d}.\:{y}'\:\:+{e}^{−\mathrm{2}{x}} {y}\:=\:\left(\mathrm{2}{x}+\mathrm{1}\right){cosx} \\ $$
Commented by abdo mathsup 649 cc last updated on 26/May/18
e.h.→y^′ + e^(−2x) y=0 ⇒y^′ =−e^(−2x) y ⇒ (y^′ /y) =−e^(−2x) ⇒ln∣y∣ = (1/2) e^(−2x) +c ⇒ y = k e^((1/2)e^(−2x) ) let use mvc method y(x)=k(x) e^((1/2)e^(−2x) ) ⇒y^′ (x)=k^′ (x)e^((1/2)e^(−2x) ) + k(x)(−e^(−2x) )e^((1/2)e^(−2x) ) (e)⇒ k^′ (x) e^((1/2)e^(−2x) ) −k(x)e^(−2x) e^((1/2)e^(−2x) ) +e^(−2x) k(x)e^((1/2)e^(−2x) ) (2x+1)cosx ⇒ k^′ (x) e^((1/2)e^(−2x) ) =(2x+1)cosx ⇒ k^′ (x) = (2x+1) e^((1/2)e^(−2x) ) cosx ⇒ k(x) = ∫ (2x+1)e^((1/2)e^(−2x) ) cosx dx +λ and y(x)={ ∫ (2x+1)e^((1/2)e^(−2x) ) cosx dx +λ} e^((1/2)e^(−2x) ) .
$${e}.{h}.\rightarrow{y}^{'} \:+\:{e}^{−\mathrm{2}{x}} \:{y}=\mathrm{0}\:\Rightarrow{y}^{'} \:=−{e}^{−\mathrm{2}{x}} \:{y}\:\Rightarrow \\ $$$$\frac{{y}^{'} }{{y}}\:=−{e}^{−\mathrm{2}{x}} \:\Rightarrow{ln}\mid{y}\mid\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−\mathrm{2}{x}} \:\:+{c}\:\Rightarrow \\ $$$$\:{y}\:=\:{k}\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } \:\:\:\:{let}\:{use}\:{mvc}\:{method}\: \\ $$$${y}\left({x}\right)={k}\left({x}\right)\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } \:\:\Rightarrow{y}^{'} \left({x}\right)={k}^{'} \left({x}\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } \: \\ $$$$+\:{k}\left({x}\right)\left(−{e}^{−\mathrm{2}{x}} \right){e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } \:\:\:\left({e}\right)\Rightarrow \\ $$$${k}^{'} \left({x}\right)\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } \:−{k}\left({x}\right){e}^{−\mathrm{2}{x}} \:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } \:+{e}^{−\mathrm{2}{x}} {k}\left({x}\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right){cosx}\:\Rightarrow\:{k}^{'} \left({x}\right)\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } \:=\left(\mathrm{2}{x}+\mathrm{1}\right){cosx}\:\Rightarrow \\ $$$${k}^{'} \left({x}\right)\:=\:\left(\mathrm{2}{x}+\mathrm{1}\right)\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } {cosx}\:\Rightarrow \\ $$$${k}\left({x}\right)\:=\:\int\:\:\:\:\left(\mathrm{2}{x}+\mathrm{1}\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } {cosx}\:{dx}\:+\lambda\:\:{and} \\ $$$${y}\left({x}\right)=\left\{\:\int\:\left(\mathrm{2}{x}+\mathrm{1}\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } {cosx}\:{dx}\:+\lambda\right\}\:{e}^{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} } . \\ $$

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