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Question Number 67014 by mathmax by abdo last updated on 21/Aug/19
solve y^(′′) +x^2 y^′ =e^(−x) sin(3x)
$${solve}\:{y}^{''} +{x}^{\mathrm{2}} {y}^{'} ={e}^{−{x}} {sin}\left(\mathrm{3}{x}\right) \\ $$
Commented by mathmax by abdo last updated on 22/Aug/19
we use the changement y^′ =z (e)→z^′ +x^2 z =e^(−x) sin(3x) (he)→z^′ +x^2 z =0 ⇒z^′ =−x^2 z ⇒(z^′ /z) =−x^2 ⇒ln∣z∣=−(x^3 /3) +c ⇒ z = K e^(−(x^3 /3)) let use mvc method z^′ =K^′ e^(−(x^3 /3)) −x^2 K e^(−(x^3 /3)) (e) ⇒(K^′ −x^2 K )e^(−(x^3 /3)) +x^2 K e^(−(x^3 /3)) =e^(−x) sin(3x) ⇒ K^′ e^(−(x^3 /3)) =e^(−x) sin(3x) ⇒K^′ =e^(−x+(x^3 /3)) sin(3x) ⇒ K(x) =∫ e^(−x+(x^3 /3)) sin(3x)dx+λ by parts u =e^((x^3 /3)−x) and v=sin(3x) K(x) =e^((x^3 /3)−x) sin(3x)−3∫ (x^2 −1)e^((x^3 /3)−x) cos(3x)dx by parts ∫ (x^2 −1)e^((x^3 /3)−x) cos(3x)dx =(1/3)(x^2 −1)e^((x^3 /3)−x) sin(3x) −(1/3)∫ { 2x e^((x^3 /3)−x) +(x^2 −1)^2 e^((x^3 /3)−1) }sin(3x)dx =.....any way let take K(x)=∫ e^((x^3 /3)−x) sin(3x)dx +λ⇒z(x) =e^(−(x^3 /3)) (∫ e^((x^3 /3)−x) sin(3x)dx+λ) =λ e^(−(x^3 /3)) + e^(−(x^3 /3)) ∫^x e^((t^3 /3)−t) sin(3t)dt y^′ =z ⇒y(x) =∫^x z(u)du +c =∫^x { λ e^(−(u^3 /3)) +e^(−(u^3 /3)) ∫^u e^((t^3 /3)−t) sin(3t)dt}du+c
$${we}\:{use}\:{the}\:{changement}\:{y}^{'} ={z}\:\:\:\:\:\:\left({e}\right)\rightarrow{z}^{'} \:+{x}^{\mathrm{2}} {z}\:={e}^{−{x}} {sin}\left(\mathrm{3}{x}\right) \\ $$$$\left({he}\right)\rightarrow{z}^{'} \:+{x}^{\mathrm{2}} {z}\:=\mathrm{0}\:\Rightarrow{z}^{'} =−{x}^{\mathrm{2}} {z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=−{x}^{\mathrm{2}} \:\Rightarrow{ln}\mid{z}\mid=−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+{c}\:\Rightarrow \\ $$$${z}\:=\:{K}\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:\:{let}\:{use}\:{mvc}\:{method}\:\:{z}^{'} ={K}^{'} \:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:−{x}^{\mathrm{2}} \:{K}\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \\ $$$$\left({e}\right)\:\Rightarrow\left({K}^{'} −{x}^{\mathrm{2}} {K}\:\right){e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:+{x}^{\mathrm{2}} {K}\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:={e}^{−{x}} {sin}\left(\mathrm{3}{x}\right)\:\Rightarrow \\ $$$${K}^{'} \:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:={e}^{−{x}} {sin}\left(\mathrm{3}{x}\right)\:\Rightarrow{K}^{'} \:={e}^{−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:{sin}\left(\mathrm{3}{x}\right)\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\int\:\:\:{e}^{−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} {sin}\left(\mathrm{3}{x}\right){dx}+\lambda\:\:{by}\:{parts}\:\:\:{u}\:={e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:{and}\:{v}={sin}\left(\mathrm{3}{x}\right) \\ $$$${K}\left({x}\right)\:={e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:{sin}\left(\mathrm{3}{x}\right)−\mathrm{3}\int\:\:\left({x}^{\mathrm{2}} −\mathrm{1}\right){e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:{cos}\left(\mathrm{3}{x}\right){dx} \\ $$$${by}\:{parts}\:\int\:\left({x}^{\mathrm{2}} −\mathrm{1}\right){e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} {cos}\left(\mathrm{3}{x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{1}\right){e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} {sin}\left(\mathrm{3}{x}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}\int\:\:\:\left\{\:\mathrm{2}{x}\:{e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:\:+\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} {e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{1}} \right\}{sin}\left(\mathrm{3}{x}\right){dx}\:=…..{any}\:{way}\:{let} \\ $$$${take}\:{K}\left({x}\right)=\int\:\:{e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:{sin}\left(\mathrm{3}{x}\right){dx}\:+\lambda\Rightarrow{z}\left({x}\right)\:={e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \left(\int\:{e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} {sin}\left(\mathrm{3}{x}\right){dx}+\lambda\right) \\ $$$$=\lambda\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:+\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:\int^{{x}} \:{e}^{\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{t}} \:{sin}\left(\mathrm{3}{t}\right){dt} \\ $$$${y}^{'} ={z}\:\Rightarrow{y}\left({x}\right)\:=\int^{{x}} {z}\left({u}\right){du}\:+{c}\:=\int^{{x}} \left\{\:\lambda\:{e}^{−\frac{{u}^{\mathrm{3}} }{\mathrm{3}}} \:+{e}^{−\frac{{u}^{\mathrm{3}} }{\mathrm{3}}} \:\int^{{u}} \:{e}^{\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{t}} \:{sin}\left(\mathrm{3}{t}\right){dt}\right\}{du}+{c} \\ $$

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