solve (x+1)y^(′′) +(2−3x^2 )y^′ =xsin(x) let y^′ =z so (e) ⇔(x+1)z^′ +(2−3x^2 )z =xsinx(e) let first find z (he) →(x+1)z^′ +(2−3x^2 )z =0 ⇒(x+1)z^′ =(3x^2 −2)z ⇒(z^′ /z) =((3x^2 −2)/(x+1)) ⇒ ln∣z∣ = ∫ ((3x^2 −2)/(x+1)) dx +c =∫ ((3(x^2 −1)+1)/(x+1))+c =∫3(x−1)dx +∫ (dx/(x+1)) +c =3((x^2 /2)−x)+ln∣x+1∣ +c ⇒z =K∣x+1∣e^((3/2)x^2 −3x) solution on]−1,+∞[ ⇒ z(x) =K(x+1) e^((3/2)x^2 −3x) mvc method give z^′ =K^′ (x+1)e^((3/2)x^2 −3x) +K{ e^((3/2)x^2 −3x) +(x+1)(3x−3) e^((3/2)x^2 −3x) } ={ K^′ (x+1) + K +3K(x+1)(x−1) } e^((3/2)x^2 −3x) (e) ⇒{(x+1)^2 K^′ +K(x+1) +3K(x+1)^2 (x−1)} e^((3/2)x^2 −3x) (2−3x^2 )K(x+1)e^((3/2)x^2 −3x) =xsinx ⇒ (x+1)K^′ +K +3K(x^2 −1) +K(2−3x^2 ) =((xsinx)/(x+1)) ⇒ (x+1)K^′ +K −K =((xsinx)/(x+1)) ⇒K^′ =((xsinx)/((x+1)^2 )) ⇒ K(x) =∫ ((xsinx)/((x+1)^2 ))dx +c_0 ⇒z(x) =(x+1)e^((3/2)x^2 −3x) { ∫ ((xsinx)/((x+1)^2 ))dx +c_0 } =c_0 (x+1) e^((3/2)x^2 −3x) +(x+1)e^((3/2)x^2 −3x) ∫_. ^x ((tsint)/((t+1)^2 )) dt we have y^′ (x)=z(x) ⇒y(x) =∫ z(x)dx +λ ⇒ y(x) =∫ c_0 (x+1)e^((3/2)x^2 −3x) dx +∫ {(x+1)e^((3/2)x^2 −3x) ∫


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Question Number 53965 by maxmathsup by imad last updated on 28/May/19
$solve (x+1)y^(′′) +(2−3x^2 )y^′ =xsin(x) let y^′ =z so (e) ⇔(x+1)z^′ +(2−3x^2 )z =xsinx(e) let first find z (he) →(x+1)z^′ +(2−3x^2 )z =0 ⇒(x+1)z^′ =(3x^2 −2)z ⇒(z^′ /z) =((3x^2 −2)/(x+1)) ⇒ ln∣z∣ = ∫ ((3x^2 −2)/(x+1)) dx +c =∫ ((3(x^2 −1)+1)/(x+1))+c =∫3(x−1)dx +∫ (dx/(x+1)) +c =3((x^2 /2)−x)+ln∣x+1∣ +c ⇒z =K∣x+1∣e^((3/2)x^2 −3x) solution on]−1,+∞[ ⇒ z(x) =K(x+1) e^((3/2)x^2 −3x) mvc method give z^′ =K^′ (x+1)e^((3/2)x^2 −3x) +K{ e^((3/2)x^2 −3x) +(x+1)(3x−3) e^((3/2)x^2 −3x) } ={ K^′ (x+1) + K +3K(x+1)(x−1) } e^((3/2)x^2 −3x) (e) ⇒{(x+1)^2 K^′ +K(x+1) +3K(x+1)^2 (x−1)} e^((3/2)x^2 −3x) (2−3x^2 )K(x+1)e^((3/2)x^2 −3x) =xsinx ⇒ (x+1)K^′ +K +3K(x^2 −1) +K(2−3x^2 ) =((xsinx)/(x+1)) ⇒ (x+1)K^′ +K −K =((xsinx)/(x+1)) ⇒K^′ =((xsinx)/((x+1)^2 )) ⇒ K(x) =∫ ((xsinx)/((x+1)^2 ))dx +c_0 ⇒z(x) =(x+1)e^((3/2)x^2 −3x) { ∫ ((xsinx)/((x+1)^2 ))dx +c_0 } =c_0 (x+1) e^((3/2)x^2 −3x) +(x+1)e^((3/2)x^2 −3x) ∫_. ^x ((tsint)/((t+1)^2 )) dt we have y^′ (x)=z(x) ⇒y(x) =∫ z(x)dx +λ ⇒ y(x) =∫ c_0 (x+1)e^((3/2)x^2 −3x) dx +∫ {(x+1)e^((3/2)x^2 −3x) ∫_. ^x ((tsint)/((t+1)^2 ))dt}dx +λ .$
$s o l v e (x + 1) y^{^{″}} + (2 - 3 x^{2}) y^{^{'}} = x s i n (x)$ $l e t y^{^{'}} = z s o (e) \Leftrightarrow (x + 1) z^{^{'}} + (2 - 3 x^{2}) z = x s i n x (e) l e t f i r s t f i n d z$ $(h e) \to (x + 1) z^{^{'}} + (2 - 3 x^{2}) z = 0 \Rightarrow (x + 1) z^{^{'}} = (3 x^{2} - 2) z \Rightarrow \frac{z^{^{'}}}{z} = \frac{3 x^{2} - 2}{x + 1} \Rightarrow$ $l n ∣ z ∣ = \int \frac{3 x^{2} - 2}{x + 1} d x + c = \int \frac{3 (x^{2} - 1) + 1}{x + 1} + c = \int 3 (x - 1) d x + \int \frac{d x}{x + 1} + c$ $= 3 (\frac{x^{2}}{2} - x) + l n ∣ x + 1 ∣ + c \Rightarrow z = K ∣ x + 1 ∣ e^{\frac{3}{2} x^{2} - 3 x} s o l u t i o n o n] - 1, + \infty [\Rightarrow$ $z (x) = K (x + 1) e^{\frac{3}{2} x^{2} - 3 x} m v c m e t h o d g i v e$ $z^{^{'}} = K^{^{'}} (x + 1) e^{\frac{3}{2} x^{2} - 3 x} + K {e^{\frac{3}{2} x^{2} - 3 x} + (x + 1) (3 x - 3) e^{\frac{3}{2} x^{2} - 3 x}}$ $= {K^{^{'}} (x + 1) + K + 3 K (x + 1) (x - 1)} e^{\frac{3}{2} x^{2} - 3 x}$ $(e) \Rightarrow {{(x + 1)}^{2} K^{^{'}} + K (x + 1) + 3 K {(x + 1)}^{2} (x - 1)} e^{\frac{3}{2} x^{2} - 3 x}$ $(2 - 3 x^{2}) K (x + 1) e^{\frac{3}{2} x^{2} - 3 x} = x s i n x \Rightarrow$ $(x + 1) K^{^{'}} + K + 3 K (x^{2} - 1) + K (2 - 3 x^{2}) = \frac{x s i n x}{x + 1} \Rightarrow$ $(x + 1) K^{^{'}} + K - K = \frac{x s i n x}{x + 1} \Rightarrow K^{^{'}} = \frac{x s i n x}{{(x + 1)}^{2}} \Rightarrow$ $K (x) = \int \frac{x s i n x}{{(x + 1)}^{2}} d x + c_{0} \Rightarrow z (x) = (x + 1) e^{\frac{3}{2} x^{2} - 3 x} {\int \frac{x s i n x}{{(x + 1)}^{2}} d x + c_{0}}$ $= c_{0} (x + 1) e^{\frac{3}{2} x^{2} - 3 x} + (x + 1) e^{\frac{3}{2} x^{2} - 3 x} \int_{.}^{x} \frac{t s i n t}{{(t + 1)}^{2}} d t$ $w e h a v e y^{^{'}} (x) = z (x) \Rightarrow y (x) = \int z (x) d x + λ \Rightarrow$ $y (x) = \int c_{0} (x + 1) e^{\frac{3}{2} x^{2} - 3 x} d x + \int {(x + 1) e^{\frac{3}{2} x^{2} - 3 x} \int_{.}^{x} \frac{t s i n t}{{(t + 1)}^{2}} d t} d x + λ .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19

	$\frac{d y}{d x} = p$ $(x + 1) \frac{d p}{d x} + (2 - 3 x^{2}) p = x s i n x$ $\frac{d p}{d x} + \frac{2 - 3 x^{2}}{x + 1} p = \frac{x s i n x}{x + 1}$ $I . F = e^{\int \frac{2 - 3 x^{2}}{x + 1} d x} = e^{\int [(- 3 x + 3) + \frac{- 1}{x + 1}] d x}$ $e^{\frac{- 3 x^{2}}{2} + 3 x - l n (x + 1)}$ $e^{\frac{- 3 x^{2}}{2} + 3 x} \times e^{- l n (x + 1)}$ $e^{\frac{- 3 x^{2}}{2} + 3 x} \times \frac{1}{x + 1}$ $e^{\frac{- 3 x^{2}}{2} + 3 x} \times \frac{1}{x + 1} \frac{d p}{d x} + \frac{2 - 3 x^{2}}{x + 1} p \times \frac{1}{x + 1} \times e^{\frac{- 3 x^{2}}{2} + 3 x} = \frac{x s i n x}{x + 1} \times e^{- \frac{3 x^{2}}{2} + 3 x} \times \frac{1}{x + 1}$ $\frac{d}{d x} (p \times \frac{e^{\frac{- 3 x^{2}}{2} + 3 x}}{x + 1}) = \frac{x s i n x \times e^{\frac{- 3 x^{2}}{2} + 3 x}}{{(x + 1)}^{2}}$ $\int d (p \times \frac{e^{- \frac{3 x^{2}}{2} + 3 x}}{x + 1}) = \int \frac{x s i n x \times e^{\frac{- 3 x^{2}}{2} + 3 x}}{{(x + 1)}^{2}} d x$ $L H S = \int d (p \times \frac{e^{- \frac{3 x^{2}}{2} + 3 x}}{x + 1}) = p \times \frac{e^{- \frac{3 x^{2}}{2} + 3 x}}{x + 1}$ $W a i t p l s . . .$
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