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


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Question Number 62438 by mathsolverby Abdo last updated on 21/Jun/19

$s p l v e x^{2} y^{^{″}} - (x + 1) y^{'} = (x + 1) e^{- x}$
Commented by mathmax by abdo last updated on 27/Jul/19
$let y^′ =z (e) ⇒x^2 z^′ −(x+1)z =(x+1)e^(−x) (he) ⇒x^2 z^′ −(x+1)z =0 ⇒x^2 z^′ =(x+1)z ⇒(z^′ /z) =((x+1)/x^2 ) =(1/x) +(1/x^2 ) ⇒ln∣z∣ =ln∣x∣−(1/x) +c ⇒z = k∣x∣e^(−(1/x)) let determine the solution on ]0,+∞[ ⇒z(x) =kx e^(−(1/x)) mvc method →z^′ =k^′ x e^(−(1/x)) +k{e^(−(1/x)) +x(1/x^2 )e^(−(1/x)) } ={k^′ x +k +(k/x)}e^(−(1/x)) (e) ⇒x^2 {k^′ x +k +(k/x)}e^(−(1/x)) −(x+1)kx e^(−(1/x)) =(x+1)e^(−x) ⇒ {k^′ x^3 +kx^2 + kx−kx^2 −kx}e^(−(1/x)) =(x+1)e^(−x) ⇒ k^′ x^3 =(x+1)e^(−x+(1/x)) ⇒k^′ =((x+1)/x^3 ) e^((1−x^2 )/x) ⇒k(x) =∫_. ^x ((t+1)/t^3 )e^((1−t^2 )/t) dt +λ ⇒z(x) =x e^(−(1/x)) { ∫_. ^x ((t+1)/t^3 )e^((1−t^2 )/t) dt +λ}=λx e^(−(1/x)) +xe^(−(1/x)) ∫_. ^x ((t+1)/t^3 )e^((1−t^2 )/t) dt y^′ (x)=z(x) ⇒y(x) =∫ z(x)dx +C y(x)=∫ λx e^(−(1/x)) dx +∫ (xe^(−(1/x)) ∫_. ^x ((t+1)/t^3 ) e^((1−t^2 )/t) dt)dx +C .$
$l e t y^{^{'}} = z (e) \Rightarrow x^{2} z^{^{'}} - (x + 1) z = (x + 1) e^{- x}$ $(h e) \Rightarrow x^{2} z^{^{'}} - (x + 1) z = 0 \Rightarrow x^{2} z^{^{'}} = (x + 1) z \Rightarrow \frac{z^{^{'}}}{z} = \frac{x + 1}{x^{2}} = \frac{1}{x} + \frac{1}{x^{2}}$ $\Rightarrow l n ∣ z ∣ = l n ∣ x ∣ - \frac{1}{x} + c \Rightarrow z = k ∣ x ∣ e^{- \frac{1}{x}}$ $l e t d e t e r m i n e t h e s o l u t i o n o n] 0, + \infty [\Rightarrow z (x) = k x e^{- \frac{1}{x}}$ $m v c m e t h o d \to z^{^{'}} = k^{^{'}} x e^{- \frac{1}{x}} + k {e^{- \frac{1}{x}} + x \frac{1}{x^{2}} e^{- \frac{1}{x}}}$ $= {k^{^{'}} x + k + \frac{k}{x}} e^{- \frac{1}{x}}$ $(e) \Rightarrow x^{2} {k^{^{'}} x + k + \frac{k}{x}} e^{- \frac{1}{x}} - (x + 1) k x e^{- \frac{1}{x}} = (x + 1) e^{- x} \Rightarrow$ ${k^{^{'}} x^{3} + {k x}^{2} + k x - {k x}^{2} - k x} e^{- \frac{1}{x}} = (x + 1) e^{- x} \Rightarrow$ $k^{^{'}} x^{3} = (x + 1) e^{- x + \frac{1}{x}} \Rightarrow k^{^{'}} = \frac{x + 1}{x^{3}} e^{\frac{1 - x^{2}}{x}} \Rightarrow k (x) = \int_{.}^{x} \frac{t + 1}{t^{3}} e^{\frac{1 - t^{2}}{t}} d t + λ$ $\Rightarrow z (x) = x e^{- \frac{1}{x}} {\int_{.}^{x} \frac{t + 1}{t^{3}} e^{\frac{1 - t^{2}}{t}} d t + λ} = λ x e^{- \frac{1}{x}} + {x e}^{- \frac{1}{x}} \int_{.}^{x} \frac{t + 1}{t^{3}} e^{\frac{1 - t^{2}}{t}} d t$ $y^{^{'}} (x) = z (x) \Rightarrow y (x) = \int z (x) d x + C$ $y (x) = \int λ x e^{- \frac{1}{x}} d x + \int ({x e}^{- \frac{1}{x}} \int_{.}^{x} \frac{t + 1}{t^{3}} e^{\frac{1 - t^{2}}{t}} d t) d x + C .$
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