solve xy^(′′) +(2+x^2 )y^′ =xe^(−x^2 )


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Question Number 98656 by mathmax by abdo last updated on 15/Jun/20

$solve {xy}^{^{″}} + (2 + x^{2}) y^{^{'}} = {xe}^{- x^{2}}$
	Answered by MWSuSon last updated on 15/Jun/20
	$let y′′=p′ and y′=p xp′+(2+x^2 )p=xe^(−x^2 ) p′+(((2+x^2 ))/x)p=e^(−x^2 ) I.F=e^(∫(((2+x^2 ))/x)dx) =e^(lnx^2 +x^2 /2) =x^2 e^(x^2 /2) x^2 e^(x^2 /2) p′+(((2+x^2 ))/x)x^2 e^(x^2 /2) p=x^2 e^(−x^2 /2) (d/dx)(x^2 e^(x^2 /2) p)=x^2 e^(−x^2 /2) x^2 e^(x^2 /2) p=∫x^2 e^(−x^2 /2) dx x^2 e^(x^2 /2) p=(√(π/2))erf((x/(√2)))−xe^(−x^2 /2) +c_1 p=(1/(x^2 e^(x^2 /2) ))[(√(π/2))erf((x/(√2)))−xe^(−x^2 /2) +c_1 ] but p=y′ y=∫{(1/(x^2 e^(x^2 /2) ))[(√(π/2))erf((x/(√2)))−xe^(−x^2 /2) +c_1 ]}dx$
	$l e t y^{″} = p^{'} a n d y^{'} = p$ ${x p}^{'} + (2 + x^{2}) p = {x e}^{- x^{2}}$ $p^{'} + \frac{(2 + x^{2})}{x} p = e^{- x^{2}}$ $I . F = e^{\int \frac{(2 + x^{2})}{x} d x} = e^{{l n x}^{2} + x^{2} / 2}$ $= x^{2} e^{x^{2} / 2}$ $x^{2} e^{x^{2} / 2} p^{'} + \frac{(2 + x^{2})}{x} x^{2} e^{x^{2} / 2} p = x^{2} e^{- x^{2} / 2}$ $\frac{d}{d x} (x^{2} e^{x^{2} / 2} p) = x^{2} e^{- x^{2} / 2}$ $x^{2} e^{x^{2} / 2} p = \int x^{2} e^{- x^{2} / 2} d x$ $x^{2} e^{x^{2} / 2} p = \sqrt{\frac{π}{2}} e r f (\frac{x}{\sqrt{2}}) - {x e}^{- x^{2} / 2} + c_{1}$ $p = \frac{1}{x^{2} e^{x^{2} / 2}} [\sqrt{\frac{π}{2}} e r f (\frac{x}{\sqrt{2}}) - {x e}^{- x^{2} / 2} + c_{1}]$ $b u t p = y^{'}$ $y = \int {\frac{1}{x^{2} e^{x^{2} / 2}} [\sqrt{\frac{π}{2}} e r f (\frac{x}{\sqrt{2}}) - {x e}^{- x^{2} / 2} + c_{1}]} d x$
	Answered by mathmax by abdo last updated on 16/Jun/20
	$let y^′ =z (e) ⇒xz^′ +(2+x^2 )z =xe^(−x^2 ) (he)→xz^′ +(x^2 +2)z =0 ⇒xz^′ =−(x^2 +2)z ⇒(z^′ /z) =−x+(2/x) ⇒ ln∣z∣ =−(x^2 /2) +2ln∣x∣ +c ⇒z =kx^2 e^(−(x^2 /2)) let use lsgrange method z^′ =k^′ x^2 e^(−(x^2 /2)) +k(2x e^(−(x^2 /2)) +x^2 (−x)e^(−(x^2 /2)) ) (e)⇒k^′ x^3 e^(−(x^2 /2)) + 2kx^2 e^(−(x^2 /2)) −kx^4 e^(−(x^2 /2)) +(x^2 +2)kx^2 e^(−(x^2 /2)) =xe^(−x^2 ) ⇒ k^′ x^3 e^(−(x^2 /2)) +4kx^2 e^(−(x^2 /2)) =xe^(−x^2 ) ⇒k^′ x^2 e^(−((x2)/2)) +4kx e^(−(x^2 /2)) =e^(−x^2 ) ⇒ x^2 k^′ +4x k =e^(−(x^2 /2)) (h)→x^2 k^′ =−4x k ⇒xk^′ =−4k ⇒(k^′ /k) =−(4/x) ⇒ln∣k∣ =−4ln∣x∣ +c ⇒ k =(α/x^4 ) (we take x>0) ⇒k^′ =(α^′ /x^4 ) +α(−4)x^(−5) e⇒(α^′ /x^2 ) −4α x^(−3) +((4α)/x^3 ) =e^(−(x^2 /2)) ⇒α^′ =x^2 e^(−(x^2 /2)) ⇒α =∫ x^2 e^(−(x^2 /2)) dx +c_0 ⇒ k =(1/x^4 ){ ∫ x^2 e^(−(x^2 /2)) dx +c_0 } ⇒z =(1/x^2 )(∫ x^2 e^(−(x^2 /2)) dx +c_0 )e^(−(x^2 /2)) y^′ =z ⇒y(x) =∫^x z(t)dt =∫^x ((e^(−(t^2 /2)) /t^2 )( ∫^t u^2 e^(−(u^2 /2)) du +c_0 )) dt+c$
	$let y^{^{'}} = z (e) \Rightarrow {xz}^{^{'}} + (2 + x^{2}) z = {xe}^{- x^{2}}$ $(he) \to {xz}^{^{'}} + (x^{2} + 2) z = 0 \Rightarrow {xz}^{^{'}} = - (x^{2} + 2) z \Rightarrow \frac{z^{^{'}}}{z} = - x + \frac{2}{x} \Rightarrow$ $\ln ∣ z ∣ = - \frac{x^{2}}{2} + 2 \ln ∣ x ∣ + c \Rightarrow z = {kx}^{2} e^{- \frac{x^{2}}{2}} let use lsgrange method$ $z^{^{'}} = k^{^{'}} x^{2} e^{- \frac{x^{2}}{2}} + k (2 x e^{- \frac{x^{2}}{2}} + x^{2} (- x) e^{- \frac{x^{2}}{2}})$ $(e) \Rightarrow k^{^{'}} x^{3} e^{- \frac{x^{2}}{2}} + {2 kx}^{2} e^{- \frac{x^{2}}{2}} - {kx}^{4} e^{- \frac{x^{2}}{2}} + (x^{2} + 2) {kx}^{2} e^{- \frac{x^{2}}{2}} = {xe}^{- x^{2}} \Rightarrow$ $k^{^{'}} x^{3} e^{- \frac{x^{2}}{2}} + {4 kx}^{2} e^{- \frac{x^{2}}{2}} = {xe}^{- x^{2}} \Rightarrow k^{^{'}} x^{2} e^{- \frac{x 2}{2}} + 4 kx e^{- \frac{x^{2}}{2}} = e^{- x^{2}} \Rightarrow$ $x^{2} k^{^{'}} + 4 x k = e^{- \frac{x^{2}}{2}}$ $(h) \to x^{2} k^{^{'}} = - 4 x k \Rightarrow {xk}^{^{'}} = - 4 k \Rightarrow \frac{k^{^{'}}}{k} = - \frac{4}{x} \Rightarrow \ln ∣ k ∣ = - 4 \ln ∣ x ∣ + c \Rightarrow$ $k = \frac{α}{x^{4}} (we take x > 0) \Rightarrow k^{^{'}} = \frac{α^{^{'}}}{x^{4}} + α (- 4) x^{- 5}$ $e \Rightarrow \frac{α^{^{'}}}{x^{2}} - 4 α x^{- 3} + \frac{4 α}{x^{3}} = e^{- \frac{x^{2}}{2}} \Rightarrow α^{^{'}} = x^{2} e^{- \frac{x^{2}}{2}} \Rightarrow α = \int x^{2} e^{- \frac{x^{2}}{2}} dx + c_{0} \Rightarrow$ $k = \frac{1}{x^{4}} {\int x^{2} e^{- \frac{x^{2}}{2}} dx + c_{0}} \Rightarrow z = \frac{1}{x^{2}} (\int x^{2} e^{- \frac{x^{2}}{2}} dx + c_{0}) e^{- \frac{x^{2}}{2}}$ $y^{^{'}} = z \Rightarrow y (x) = \int^{x} z (t) dt = \int^{x} (\frac{e^{- \frac{t^{2}}{2}}}{t^{2}} (\int^{t} u^{2} e^{- \frac{u^{2}}{2}} du + c_{0})) dt + c$
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