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Question Number 63894 by mathmax by abdo last updated on 10/Jul/19

sove the (de) x^2 y^′ −(2x+3)y =sin(x^2 ) with y(1)=2 and y^′ (1)=1 .

$${sove}\:{the}\:\left({de}\right)\:{x}^{\mathrm{2}} {y}^{'} \:−\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:={sin}\left({x}^{\mathrm{2}} \right)\:\:{with}\:{y}\left(\mathrm{1}\right)=\mathrm{2}\:{and} \\ $$$${y}^{'} \left(\mathrm{1}\right)=\mathrm{1}\:. \\ $$

Commented by mathmax by abdo last updated on 11/Jul/19

(he) →x^2 y^′ −(2x+3)y=0 ⇒x^2 y^′ =(2x+3)y ⇒ (y^′ /y) =((2x+3)/x^2 ) =(2/x) +(3/x^2 ) ⇒ln∣y∣=2ln∣x∣−(3/x)+c ⇒ y =K x^2 e^(−(3/x)) let use mvc method y^′ =K^′ x^2 e^(−(3/x)) +K{ 2x e^(−(3/x)) +x^2 (3/x^2 ) e^(−(3/x)) } =K^′ x^2 e^(−(3/(x ))) +K{2x+3}e^(−(3/x)) (e) ⇒K^′ x^4 e^(−(3/x)) +Kx^2 (2x+3)e^(−(3/x)) −(2x+3)Kx^2 e^(−(3/x)) =sin(x^2 ) ⇒ K^′ x^4 e^(−(3/x)) =sin(x^2 ) ⇒K^′ =((sin(x^2 )e^(3/x) )/x^4 ) ⇒ K(x) =∫_1 ^x ((sin(t^2 )e^(3/t) )/t^4 )dt +λ λ=K(1) we have y(1)=K(1)e^(−3) ⇒K(1)=e^3 y(1) ⇒ K(x) =∫_1 ^x ((sin(t^2 )e^(3/t) )/t^4 )dt +2e^3 ⇒ y(x) =x^2 e^(−(3/x)) { ∫_1 ^x ((e^(3/t) sin(t^2 ))/t^4 )dt +2e^3 }....

$$\left({he}\right)\:\rightarrow{x}^{\mathrm{2}} {y}^{'} −\left(\mathrm{2}{x}+\mathrm{3}\right){y}=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} {y}^{'} \:=\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:\Rightarrow \\ $$$$\frac{{y}^{'} }{{y}}\:=\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{{x}}\:+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:\Rightarrow{ln}\mid{y}\mid=\mathrm{2}{ln}\mid{x}\mid−\frac{\mathrm{3}}{{x}}+{c}\:\Rightarrow \\ $$$${y}\:={K}\:{x}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:\:\:{let}\:{use}\:{mvc}\:{method}\: \\ $$$${y}^{'} \:={K}^{'} {x}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:+{K}\left\{\:\mathrm{2}{x}\:{e}^{−\frac{\mathrm{3}}{{x}}} \:+{x}^{\mathrm{2}} \frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:{e}^{−\frac{\mathrm{3}}{{x}}} \right\} \\ $$$$={K}^{'} \:{x}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}\:}} \:+{K}\left\{\mathrm{2}{x}+\mathrm{3}\right\}{e}^{−\frac{\mathrm{3}}{{x}}} \\ $$$$\left({e}\right)\:\Rightarrow{K}^{'} {x}^{\mathrm{4}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:\:+{Kx}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{3}\right){e}^{−\frac{\mathrm{3}}{{x}}} \:−\left(\mathrm{2}{x}+\mathrm{3}\right){Kx}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:={sin}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${K}^{'} \:{x}^{\mathrm{4}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:={sin}\left({x}^{\mathrm{2}} \right)\:\Rightarrow{K}^{'} \:=\frac{{sin}\left({x}^{\mathrm{2}} \right){e}^{\frac{\mathrm{3}}{{x}}} }{{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\int_{\mathrm{1}} ^{{x}} \:\:\:\frac{{sin}\left({t}^{\mathrm{2}} \right){e}^{\frac{\mathrm{3}}{{t}}} }{{t}^{\mathrm{4}} }{dt}\:\:+\lambda \\ $$$$\lambda={K}\left(\mathrm{1}\right)\:{we}\:{have}\:{y}\left(\mathrm{1}\right)={K}\left(\mathrm{1}\right){e}^{−\mathrm{3}} \:\Rightarrow{K}\left(\mathrm{1}\right)={e}^{\mathrm{3}} {y}\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\int_{\mathrm{1}} ^{{x}} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right){e}^{\frac{\mathrm{3}}{{t}}} }{{t}^{\mathrm{4}} }{dt}\:\:+\mathrm{2}{e}^{\mathrm{3}} \:\Rightarrow \\ $$$${y}\left({x}\right)\:={x}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \left\{\:\int_{\mathrm{1}} ^{{x}} \:\:\frac{{e}^{\frac{\mathrm{3}}{{t}}} \:{sin}\left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{4}} }{dt}\:+\mathrm{2}{e}^{\mathrm{3}} \right\}.... \\ $$

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