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Question Number 46857 by maxmathsup by imad last updated on 01/Nov/18
solve 2x y^′ +(1+x^2 )y =xe^(−x) withy(o)=1
$${solve}\:\mathrm{2}{x}\:{y}^{'} \:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}\:={xe}^{−{x}} \:\:\:{withy}\left({o}\right)=\mathrm{1}\: \\ $$
Commented by maxmathsup by imad last updated on 11/Nov/18
(he) ⇒2xy^′ +(1+x^2 )y =0 ⇒2xy^′ =−(1+x^2 )y ⇒(y^′ /y) =−((1+x^2 )/(2x)) =−(1/(2x)) −(x/2) ⇒∫ (y^′ /y)dx =−(1/2)ln∣x∣−(x^2 /4)+k ⇒ln∣y∣=ln((1/( (√(∣x∣))))) −(x^2 /4) +k ⇒ y =K (e^(−(x^2 /4)) /( (√(∣x∣)))) let suppose x>0 ⇒y=K x^(−(1/2)) e^(−(x^2 /4)) mvc method give y^′ =K^′ x^(−(1/2)) e^(−(x^2 /4)) +K( −(1/2)x^(−(3/2)) e^(−(x^2 /4)) −(x/2) x^(−(1/2)) e^(−(x^2 /4)) ) =(K^′ x^(−(1/2)) −(K/2) x^(−(3/2)) −(((√x)K)/2) )e^(−(x^2 /4)) (e) ⇔ 2x(K^′ x^(−(1/2)) −(K/2) x^(−(3/2)) −((K(√x))/2))e^(−(x^2 /4)) +(1+x^2 )Kx^(−(1/2)) e^(−(x^2 /4)) =xe^(−x) ⇒ 2(√x)K^′ −Kx^(−(1/2)) −Kx^(3/2) +K x^(−(1/2)) +K x^(3/2) =xe^(−x) e^(x^2 /4) ⇒ K^′ =((xe^(−x) )/(2(√x))) e^(x^2 /4) ⇒K^′ =(((√x)e^(−x) )/2) e^(x^2 /4) ⇒ K(x)=∫_0 ^x (√t)e^(−t+(t^2 /4)) dt+λ ⇒ y(x)=(e^(−(x^2 /4)) /( (√x))) { ∫_0 ^x (√t)e^(−t+(t^2 /4)) dt +λ} y(1)=0 ⇒e^(−(1/4)) { ∫_0 ^1 (√t)e^((t^2 /4)−t) dt +λ}=0 ⇒λ=−∫_0 ^1 (√t)e^((t^2 /4)−t) dt ⇒ y(x)=(e^(−(x^2 /4)) /( (√x))) ∫_1 ^x (√t)e^((t^2 /4)−t) dt .
$$\left({he}\right)\:\Rightarrow\mathrm{2}{xy}^{'} \:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}\:=\mathrm{0}\:\Rightarrow\mathrm{2}{xy}^{'} =−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}\:\Rightarrow\frac{{y}^{'} }{{y}}\:=−\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{x}}\:−\frac{{x}}{\mathrm{2}}\:\Rightarrow\int\:\frac{{y}^{'} }{{y}}{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}\mid−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+{k}\:\Rightarrow{ln}\mid{y}\mid={ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mid{x}\mid}}\right)\:−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:+{k}\:\Rightarrow \\ $$$${y}\:={K}\:\frac{{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} }{\:\sqrt{\mid{x}\mid}}\:\:{let}\:{suppose}\:{x}>\mathrm{0}\:\Rightarrow{y}={K}\:{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\:{mvc}\:{method}\:{give} \\ $$$${y}^{'} \:={K}^{'} \:{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:+{K}\left(\:−\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\frac{\mathrm{3}}{\mathrm{2}}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:−\frac{{x}}{\mathrm{2}}\:{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \right) \\ $$$$=\left({K}^{'} \:{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} −\frac{{K}}{\mathrm{2}}\:{x}^{−\frac{\mathrm{3}}{\mathrm{2}}} \:−\frac{\sqrt{{x}}{K}}{\mathrm{2}}\:\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\left({e}\right)\:\Leftrightarrow\:\mathrm{2}{x}\left({K}^{'} \:{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:−\frac{{K}}{\mathrm{2}}\:{x}^{−\frac{\mathrm{3}}{\mathrm{2}}} \:−\frac{{K}\sqrt{{x}}}{\mathrm{2}}\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){Kx}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:={xe}^{−{x}} \:\Rightarrow \\ $$$$\mathrm{2}\sqrt{{x}}{K}^{'} \:−{Kx}^{−\frac{\mathrm{1}}{\mathrm{2}}} −{Kx}^{\frac{\mathrm{3}}{\mathrm{2}}} \:+{K}\:{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\:+{K}\:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:={xe}^{−{x}} \:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\Rightarrow \\ $$$${K}^{'} \:=\frac{{xe}^{−{x}} }{\mathrm{2}\sqrt{{x}}}\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\Rightarrow{K}^{'} \:=\frac{\sqrt{{x}}{e}^{−{x}} }{\mathrm{2}}\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:\Rightarrow\:{K}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\sqrt{{t}}{e}^{−{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{4}}} {dt}+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)=\frac{{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} }{\:\sqrt{{x}}}\:\left\{\:\int_{\mathrm{0}} ^{{x}} \sqrt{{t}}{e}^{−{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{4}}} {dt}\:+\lambda\right\} \\ $$$${y}\left(\mathrm{1}\right)=\mathrm{0}\:\Rightarrow{e}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\left\{\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{{t}}{e}^{\frac{{t}^{\mathrm{2}} }{\mathrm{4}}−{t}} {dt}\:+\lambda\right\}=\mathrm{0}\:\Rightarrow\lambda=−\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{t}}{e}^{\frac{{t}^{\mathrm{2}} }{\mathrm{4}}−{t}} {dt}\:\:\Rightarrow \\ $$$${y}\left({x}\right)=\frac{{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} }{\:\sqrt{{x}}}\:\int_{\mathrm{1}} ^{{x}} \sqrt{{t}}{e}^{\frac{{t}^{\mathrm{2}} }{\mathrm{4}}−{t}} {dt}\:. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 11/Nov/18
sorry the condition is y(1)=0.
$${sorry}\:{the}\:{condition}\:{is}\:{y}\left(\mathrm{1}\right)=\mathrm{0}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18
2x(dy/dx)+(1+x^2 )y=xe^(−x) (dy/dx)+((1+x^2 )/(2x)).y=(e^(−x) /2) e^(∫((1/(2x))+(x/2))dx ←intregating factor) e^((1/2)lnx+(x^2 /4)) e^(ln(√x) ) ×e^(x^2 /4) (√x) ×e^(x^2 /4) (√x) ×e^(x^2 /4) ×(dy/dx)+(1+x^2 )×(√x) e^(x^2 /4) ×((1+x^2 )/(2x))=(√x) ×e^(x^2 /4) ×(e^(−x) /2) wait ...busy..
$$\mathrm{2}{x}\frac{{dy}}{{dx}}+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}={xe}^{−{x}} \\ $$$$\frac{{dy}}{{dx}}+\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}}.{y}=\frac{{e}^{−{x}} }{\mathrm{2}} \\ $$$${e}^{\int\left(\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{{x}}{\mathrm{2}}\right){dx}\:\:\:\leftarrow{intregating}\:{factor}} \\ $$$${e}^{\frac{\mathrm{1}}{\mathrm{2}}{lnx}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${e}^{{ln}\sqrt{{x}}\:} ×{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\sqrt{{x}}\:×{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\sqrt{{x}}\:×{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} ×\frac{{dy}}{{dx}}+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)×\sqrt{{x}}\:{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} ×\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}{x}}=\sqrt{{x}}\:×{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} ×\frac{{e}^{−{x}} }{\mathrm{2}} \\ $$$${wait}\:…{busy}.. \\ $$$$ \\ $$$$ \\ $$

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