solve x^2 y^(′′) −(x+1)y^′ =x^2 sinx


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Question Number 97625 by mathmax by abdo last updated on 08/Jun/20

$solve x^{2} y^{^{″}} - (x + 1) y^{^{'}} = x^{2} sinx$
	Answered by mathmax by abdo last updated on 09/Jun/20
	$let y^′ =z so (e)⇒x^2 z^′ −(x+1)z =x^2 sinx (he)→x^2 z^′ −(x+1)z =0 ⇒x^2 z^′ =(x+1)z ⇒(z^′ /z) =((x+1)/x^2 ) =(1/x) +(1/x^2 ) ⇒ ln∣z∣ =ln∣x∣−(1/x) +c ⇒z =k∣x∣e^(−(1/x)) let find solution on ]0,+∞[ ⇒ z =kxe^(−(1/x)) mvc method →z^′ =k^′ x e^(−(1/x)) +k(e^(−(1/x)) +x×(1/x^2 )e^(−(1/x)) ) =k^′ x e^(−(1/x)) +k e^(−(1/x)) +(k/x) e^(−(1/x)) (e)⇒k^′ x^3 e^(−(1/x)) +kx^2 e^(−(1/x)) +kx e^(−(1/x)) −kx(x+1)e^(−(1/x)) =x^2 sinx ⇒ (k^′ x^3 +kx^2 +k−kx^2 −kx)e^(−(1/x)) =x^2 sinx ⇒k^′ x^3 e^(−(1/x)) =x^2 sinx ⇒ k^′ =((sinx)/x) e^(1/x) ⇒ k(x) = ∫^x ((sint)/t) e^(1/t) dt +c ⇒ z(x) =x e^(−(1/x)) { ∫^x ((sint)/t)e^(1/t) dt +c} =x e^(−(1/x)) ∫^x ((sint)/t)e^(1/t) dt +cx e^(−(1/x)) y^′ =z ⇒y(x) =∫^x z(u)du +λ =∫^x {u e^(−(1/u)) ∫^u ((sint)/t)e^(1/t) dt +cu e^(−(1/u)) }du +λ$
	$let y^{^{'}} = z so (e) \Rightarrow x^{2} z^{^{'}} - (x + 1) z = x^{2} sinx$ $(he) \to x^{2} z^{^{'}} - (x + 1) z = 0 \Rightarrow x^{2} z^{^{'}} = (x + 1) z \Rightarrow \frac{z^{^{'}}}{z} = \frac{x + 1}{x^{2}} = \frac{1}{x} + \frac{1}{x^{2}} \Rightarrow$ $\ln ∣ z ∣ = \ln ∣ x ∣ - \frac{1}{x} + c \Rightarrow z = k ∣ x ∣ e^{- \frac{1}{x}} let find solution on] 0, + \infty [\Rightarrow$ $z = {kxe}^{- \frac{1}{x}} mvc method \to z^{^{'}} = k^{^{'}} x e^{- \frac{1}{x}} + k (e^{- \frac{1}{x}} + x \times \frac{1}{x^{2}} e^{- \frac{1}{x}})$ $= k^{^{'}} x e^{- \frac{1}{x}} + k e^{- \frac{1}{x}} + \frac{k}{x} e^{- \frac{1}{x}}$ $(e) \Rightarrow k^{^{'}} x^{3} e^{- \frac{1}{x}} + {kx}^{2} e^{- \frac{1}{x}} + kx e^{- \frac{1}{x}} - kx (x + 1) e^{- \frac{1}{x}} = x^{2} sinx \Rightarrow$ $(k^{^{'}} x^{3} + {kx}^{2} + k - {kx}^{2} - kx) e^{- \frac{1}{x}} = x^{2} sinx \Rightarrow k^{^{'}} x^{3} e^{- \frac{1}{x}} = x^{2} sinx \Rightarrow$ $k^{^{'}} = \frac{sinx}{x} e^{\frac{1}{x}} \Rightarrow k (x) = \int^{x} \frac{sint}{t} e^{\frac{1}{t}} dt + c \Rightarrow$ $z (x) = x e^{- \frac{1}{x}} {\int^{x} \frac{sint}{t} e^{\frac{1}{t}} dt + c} = x e^{- \frac{1}{x}} \int^{x} \frac{sint}{t} e^{\frac{1}{t}} dt + cx e^{- \frac{1}{x}}$ $y^{^{'}} = z \Rightarrow y (x) = \int^{x} z (u) du + λ$ $= \int^{x} {u e^{- \frac{1}{u}} \int^{u} \frac{sint}{t} e^{\frac{1}{t}} dt + cu e^{- \frac{1}{u}}} du + λ$
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