solve 2x y^′ +(1+x^2 )y =xe^(−x) withy(o)=1


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Question Number 46857 by maxmathsup by imad last updated on 01/Nov/18

$s o l v e 2 x y^{^{'}} + (1 + x^{2}) y = {x e}^{- x} w i t h y (o) = 1$
Commented by maxmathsup by imad last updated on 11/Nov/18
$(he) ⇒2xy^′ +(1+x^2 )y =0 ⇒2xy^′ =−(1+x^2 )y ⇒(y^′ /y) =−((1+x^2 )/(2x)) =−(1/(2x)) −(x/2) ⇒∫ (y^′ /y)dx =−(1/2)ln∣x∣−(x^2 /4)+k ⇒ln∣y∣=ln((1/(√(∣x∣)))) −(x^2 /4) +k ⇒ y =K (e^(−(x^2 /4)) /(√(∣x∣))) let suppose x>0 ⇒y=K x^(−(1/2)) e^(−(x^2 /4)) mvc method give y^′ =K^′ x^(−(1/2)) e^(−(x^2 /4)) +K( −(1/2)x^(−(3/2)) e^(−(x^2 /4)) −(x/2) x^(−(1/2)) e^(−(x^2 /4)) ) =(K^′ x^(−(1/2)) −(K/2) x^(−(3/2)) −(((√x)K)/2) )e^(−(x^2 /4)) (e) ⇔ 2x(K^′ x^(−(1/2)) −(K/2) x^(−(3/2)) −((K(√x))/2))e^(−(x^2 /4)) +(1+x^2 )Kx^(−(1/2)) e^(−(x^2 /4)) =xe^(−x) ⇒ 2(√x)K^′ −Kx^(−(1/2)) −Kx^(3/2) +K x^(−(1/2)) +K x^(3/2) =xe^(−x) e^(x^2 /4) ⇒ K^′ =((xe^(−x) )/(2(√x))) e^(x^2 /4) ⇒K^′ =(((√x)e^(−x) )/2) e^(x^2 /4) ⇒ K(x)=∫_0 ^x (√t)e^(−t+(t^2 /4)) dt+λ ⇒ y(x)=(e^(−(x^2 /4)) /(√x)) { ∫_0 ^x (√t)e^(−t+(t^2 /4)) dt +λ} y(1)=0 ⇒e^(−(1/4)) { ∫_0 ^1 (√t)e^((t^2 /4)−t) dt +λ}=0 ⇒λ=−∫_0 ^1 (√t)e^((t^2 /4)−t) dt ⇒ y(x)=(e^(−(x^2 /4)) /(√x)) ∫_1 ^x (√t)e^((t^2 /4)−t) dt .$
$(h e) \Rightarrow 2 {x y}^{^{'}} + (1 + x^{2}) y = 0 \Rightarrow 2 {x y}^{^{'}} = - (1 + x^{2}) y \Rightarrow \frac{y^{^{'}}}{y} = - \frac{1 + x^{2}}{2 x}$ $= - \frac{1}{2 x} - \frac{x}{2} \Rightarrow \int \frac{y^{^{'}}}{y} d x = - \frac{1}{2} l n ∣ x ∣ - \frac{x^{2}}{4} + k \Rightarrow l n ∣ y ∣= l n (\frac{1}{\sqrt{∣ x ∣}}) - \frac{x^{2}}{4} + k \Rightarrow$ $y = K \frac{e^{- \frac{x^{2}}{4}}}{\sqrt{∣ x ∣}} l e t s u p p o s e x > 0 \Rightarrow y = K x^{- \frac{1}{2}} e^{- \frac{x^{2}}{4}} m v c m e t h o d g i v e$ $y^{^{'}} = K^{^{'}} x^{- \frac{1}{2}} e^{- \frac{x^{2}}{4}} + K (- \frac{1}{2} x^{- \frac{3}{2}} e^{- \frac{x^{2}}{4}} - \frac{x}{2} x^{- \frac{1}{2}} e^{- \frac{x^{2}}{4}})$ $= (K^{^{'}} x^{- \frac{1}{2}} - \frac{K}{2} x^{- \frac{3}{2}} - \frac{\sqrt{x} K}{2}) e^{- \frac{x^{2}}{4}}$ $(e) \Leftrightarrow 2 x (K^{^{'}} x^{- \frac{1}{2}} - \frac{K}{2} x^{- \frac{3}{2}} - \frac{K \sqrt{x}}{2}) e^{- \frac{x^{2}}{4}} + (1 + x^{2}) {K x}^{- \frac{1}{2}} e^{- \frac{x^{2}}{4}} = {x e}^{- x} \Rightarrow$ $2 \sqrt{x} K^{^{'}} - {K x}^{- \frac{1}{2}} - {K x}^{\frac{3}{2}} + K x^{- \frac{1}{2}} + K x^{\frac{3}{2}} = {x e}^{- x} e^{\frac{x^{2}}{4}} \Rightarrow$ $K^{^{'}} = \frac{{x e}^{- x}}{2 \sqrt{x}} e^{\frac{x^{2}}{4}} \Rightarrow K^{^{'}} = \frac{\sqrt{x} e^{- x}}{2} e^{\frac{x^{2}}{4}} \Rightarrow K (x) = \int_{0}^{x} \sqrt{t} e^{- t + \frac{t^{2}}{4}} d t + λ \Rightarrow$ $y (x) = \frac{e^{- \frac{x^{2}}{4}}}{\sqrt{x}} {\int_{0}^{x} \sqrt{t} e^{- t + \frac{t^{2}}{4}} d t + λ}$ $y (1) = 0 \Rightarrow e^{- \frac{1}{4}} {\int_{0}^{1} \sqrt{t} e^{\frac{t^{2}}{4} - t} d t + λ} = 0 \Rightarrow λ = - \int_{0}^{1} \sqrt{t} e^{\frac{t^{2}}{4} - t} d t \Rightarrow$ $y (x) = \frac{e^{- \frac{x^{2}}{4}}}{\sqrt{x}} \int_{1}^{x} \sqrt{t} e^{\frac{t^{2}}{4} - t} d t .$
Commented by maxmathsup by imad last updated on 11/Nov/18

$s o r r y t h e c o n d i t i o n i s y (1) = 0 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18

	$2 x \frac{d y}{d x} + (1 + x^{2}) y = {x e}^{- x}$ $\frac{d y}{d x} + \frac{1 + x^{2}}{2 x} . y = \frac{e^{- x}}{2}$ $e^{\int (\frac{1}{2 x} + \frac{x}{2}) d x \leftarrow i n t r e g a t i n g f a c t o r}$ $e^{\frac{1}{2} l n x + \frac{x^{2}}{4}}$ $e^{l n \sqrt{x}} \times e^{\frac{x^{2}}{4}}$ $\sqrt{x} \times e^{\frac{x^{2}}{4}}$ $\sqrt{x} \times e^{\frac{x^{2}}{4}} \times \frac{d y}{d x} + (1 + x^{2}) \times \sqrt{x} e^{\frac{x^{2}}{4}} \times \frac{1 + x^{2}}{2 x} = \sqrt{x} \times e^{\frac{x^{2}}{4}} \times \frac{e^{- x}}{2}$ $w a i t . . . b u s y . .$
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