sove (x^2 −3x)y^(′′) +2x y^′ =(2x+1)e^(−x^2 )


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Question Number 69790 by mathmax by abdo last updated on 27/Sep/19

$s o v e (x^{2} - 3 x) y^{^{″}} + 2 x y^{^{'}} = (2 x + 1) e^{- x^{2}}$
Commented by mathmax by abdo last updated on 01/Oct/19
$we put y^′ =z so (e) ⇔(x^2 −3x)z^′ +2xz =(2x+1)e^(−x^2 ) (e) (he) →(x^2 −3x)z^′ +2x z =0 ⇒(x^2 −3x)z^′ =−2xz ⇒ (z^′ /z)=((−2x)/(x^2 −3x)) =((−2)/(x−3)) ⇒ln∣z∣=−2ln∣x−3∣ +α ⇒ z =(K/(∣x−3∣^2 )) let determine the solution on ]3,+∞[ z =(K/((x−3)^2 )) mvc method →z^′ =(K^′ /((x−3)^2 )) +K×((−2(x−3)^ )/((x−3)^4 )) =(K^′ /((x−3)^2 )) −((2K)/((x−3)^3 )) (e) ⇒(x^2 −3x){(K^′ /((x−3)^2 ))−((2K)/((x−3)^3 ))}+2x(K/((x−3)^2 )) =(2x+1)e^(−x^2 ) ⇒ (x/(x−3))K^′ −2(x/((x−3)^2 ))K +((2xK)/((x−3)^2 )) =(2x+1)e^(−x^2 ) ⇒ K^′ =(((2x+1)(x−3))/x) e^(−x^2 ) ⇒K^′ =((2x^2 −6x+x−3)/x)e^(−x^2 ) ⇒ K^′ =(((2x^2 −5x−3))/x) e^(−x^2 ) ⇒K(x) =∫^x (((2t^2 −5t−3))/t) e^(−t^2 ) dt +C ⇒ z(x)=(1/((x−3)^2 )){ ∫^x (((2t^2 −5t−3))/t)e^(−t^2 ) dt +C} y^′ (x)=z(x) ⇒y(x) =∫z(x)dx z is known$
$w e p u t y^{^{'}} = z s o (e) \Leftrightarrow (x^{2} - 3 x) z^{^{'}} + 2 x z = (2 x + 1) e^{- x^{2}} (e)$ $(h e) \to (x^{2} - 3 x) z^{^{'}} + 2 x z = 0 \Rightarrow (x^{2} - 3 x) z^{^{'}} = - 2 x z \Rightarrow$ $\frac{z^{^{'}}}{z} = \frac{- 2 x}{x^{2} - 3 x} = \frac{- 2}{x - 3} \Rightarrow l n ∣ z ∣= - 2 l n ∣ x - 3 ∣ + α \Rightarrow$ $z = \frac{K}{∣ x - 3 ∣^{2}} l e t d e t e r m i n e t h e s o l u t i o n o n] 3, + \infty [$ $z = \frac{K}{{(x - 3)}^{2}} m v c m e t h o d \to z^{^{'}} = \frac{K^{^{'}}}{{(x - 3)}^{2}} + K \times \frac{- 2 {(x - 3)}^{}}{{(x - 3)}^{4}}$ $= \frac{K^{^{'}}}{{(x - 3)}^{2}} - \frac{2 K}{{(x - 3)}^{3}}$ $(e) \Rightarrow (x^{2} - 3 x) {\frac{K^{^{'}}}{{(x - 3)}^{2}} - \frac{2 K}{{(x - 3)}^{3}}} + 2 x \frac{K}{{(x - 3)}^{2}} = (2 x + 1) e^{- x^{2}} \Rightarrow$ $\frac{x}{x - 3} K^{^{'}} - 2 \frac{x}{{(x - 3)}^{2}} K + \frac{2 x K}{{(x - 3)}^{2}} = (2 x + 1) e^{- x^{2}} \Rightarrow$ $K^{^{'}} = \frac{(2 x + 1) (x - 3)}{x} e^{- x^{2}} \Rightarrow K^{^{'}} = \frac{2 x^{2} - 6 x + x - 3}{x} e^{- x^{2}} \Rightarrow$ $K^{^{'}} = \frac{(2 x^{2} - 5 x - 3)}{x} e^{- x^{2}} \Rightarrow K (x) = \int^{x} \frac{(2 t^{2} - 5 t - 3)}{t} e^{- t^{2}} d t + C \Rightarrow$ $z (x) = \frac{1}{{(x - 3)}^{2}} {\int^{x} \frac{(2 t^{2} - 5 t - 3)}{t} e^{- t^{2}} d t + C}$ $y^{^{'}} (x) = z (x) \Rightarrow y (x) = \int z (x) d x z i s k n o w n$
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