(x^2 +x) (d^2 y/dx^2 ) + (1−2x^2 ) (dy/dx) + (x^2 −x−1)y = x^2 −x−1


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Question Number 97847 by bemath last updated on 10/Jun/20

$(x^{2} + x) \frac{d^{2} y}{{dx}^{2}} + (1 - {2 x}^{2}) \frac{dy}{dx} + (x^{2} - x - 1) y = x^{2} - x - 1$
	Answered by niroj last updated on 10/Jun/20
	$(x^2 +x)(d^2 y/dx^2 ) + (1−2x^2 ) (dy/dx) +(x^2 −x−1)y= x^2 −x−1 (d^2 y/dx^2 ) + (((1−2x^2 ))/(x^2 +x)) (dy/dx) + ((x^2 −x−1)/(x^2 +x)) y = ((x^2 −x−1)/(x^2 +x)) We know variable coefficient for CF If : 1+P+Q=0 then part of CF ,u=e^x let P= ((1−2x^2 )/(x^2 +x)) , Q= ((x^2 −x−1)/(x^2 +x)) and R= ((x^2 −x−1)/(x^2 +x)) 1+((1−2x^2 )/(x^2 +x))+((x^2 −x−1)/(x^2 +x))= ((x^2 +x+1−2x^2 +x^2 −x−1)/(x^2 +x))= (0/(x^2 +x))=0. ∵1+P+Q=0 is satisfy complete function y= uv y= e^x v.......(i) we know special case : (d^2 v/dx^2 )+(p+(2/u).(du/dx))(dv/dx)= (R/u) (d^2 v/dx^2 )+(((1−2x^2 )/(x^2 +x)) + (2/e^x ) .e^x )(dv/dx)= ((x^2 −x−1)/(e^x (x^2 +x))) (d^2 v/dx^2 ) +(((2x^2 +2x+1−2x^2 )/(x^2 +x)))(dv/dx) = ((x^2 −x−1)/(e^x (x^2 +x))) (d^2 v/dx^2 ) + ((2x+1)/(x^2 +x)) (dv/dx) = ((x^2 −x−1)/(e^x (x^2 +x))) Now we reduces linear form, Put , (dv/dx)= t ⇒ (d^2 v/dx^2 )= (dt/dx) (dt/dx) + ((2x+1)/(x^2 +x )). t = ((x^2 −x−1)/(e^x (x^2 +x))) IF= e^(∫Pdx) = e^(∫((2x+1)/(x^2 +x))dx) =e^(log (x^2 +x)) IF= x^2 +x t.IF= ∫IF.Qdx+C_1 t.(x^2 +x)= ∫(x^2 +x).((x^2 −x−1)/(e^x (x^2 +x)))dx+C_1 t(x^2 +x)= ∫ ((x^2 −x−1)/e^x )dx+C_1 t(x^2 +x)= ∫ x^2 e^(−x) dx−∫xe^(−x) dx−∫e^(−x) dx+C_1 t(x^2 +x)= [x^2 .(−e^(−x) )−∫2x.(−e^(−x) )dx]−∫xe^(−x) dx−(−e^(−x) )+C_1 t(x^2 +x)= −x^2 e^(−x) +2∫xe^(−x) dx−∫xe^(−x) dx+e^(−x) +C_1 t(x^2 +x)= −x^2 e^(−x) +∫xe^(−x) dx+e^(−x) +C_1 t(x^2 +x)= −x^2 e^(−x) +x(−e^x )−∫1.(−e^(−x) )dx+e^(−x) +C_1 t(x^2 +x)= −x^2 e^(−x) −xe^(−x) −e^(−x) +e^(−x) +C_1 t(x^2 +x)= −x^2 e^(−x) −xe^(−x) +C_1 t= ((−xe^(−x) (x+1))/(x(x+1)))+C_1 (1/(x(x+1))) (dv/dx) = −e^(−x) +C_1 (1/(x(x+1))) ∫dv = −∫e^(−x) dx+C_1 ∫ (1/(x(x+1)))dx+C_2 v= −(−e^(−x) )+C_1 [ ∫(1/x)dx−∫(1/(x+1))dx]+C_2 v= e^(−x) +C_1 { log x−log(x+1)}+C_2 v= e^(−x) +C_1 log (x/(x+1))+C_2 Now, again put value of v in eq^n (i) y=e^x v.....(i) y= e^x (e^(−x) +C_1 log (x/(x+1))+C_2 ) y = 1+ C_1 e^x log (x/(x+1)) +C_(2 ) e^x //.$
	$(x^{2} + x) \frac{d^{2} y}{{dx}^{2}} + (1 - {2 x}^{2}) \frac{dy}{dx} + (x^{2} - x - 1) y = x^{2} - x - 1$ $\frac{d^{2} y}{{dx}^{2}} + \frac{(1 - {2 x}^{2})}{x^{2} + x} \frac{dy}{dx} + \frac{x^{2} - x - 1}{x^{2} + x} y = \frac{x^{2} - x - 1}{x^{2} + x}$ $We know variable coefficient for CF$ $If : 1 + P + Q = 0 then part of CF, u = e^{x}$ $let P = \frac{1 - {2 x}^{2}}{x^{2} + x}, Q = \frac{x^{2} - x - 1}{x^{2} + x} and R = \frac{x^{2} - x - 1}{x^{2} + x}$ $1 + \frac{1 - {2 x}^{2}}{x^{2} + x} + \frac{x^{2} - x - 1}{x^{2} + x} = \frac{x^{2} + x + 1 - {2 x}^{2} + x^{2} - x - 1}{x^{2} + x} = \frac{0}{x^{2} + x} = 0 .$ $∵ 1 + P + Q = 0 is satisfy$ $complete function$ $y = uv$ $y = e^{x} v . . . . . . . (i)$ $we know special case :$ $\frac{d^{2} v}{{dx}^{2}} + (p + \frac{2}{u} . \frac{du}{dx}) \frac{dv}{dx} = \frac{R}{u}$ $\frac{d^{2} v}{{dx}^{2}} + (\frac{1 - {2 x}^{2}}{x^{2} + x} + \frac{2}{e^{x}} . e^{x}) \frac{dv}{dx} = \frac{x^{2} - x - 1}{e^{x} (x^{2} + x)}$ $\frac{d^{2} v}{{dx}^{2}} + (\frac{{2 x}^{2} + 2 x + 1 - {2 x}^{2}}{x^{2} + x}) \frac{dv}{dx} = \frac{x^{2} - x - 1}{e^{x} (x^{2} + x)}$ $\frac{d^{2} v}{{dx}^{2}} + \frac{2 x + 1}{x^{2} + x} \frac{dv}{dx} = \frac{x^{2} - x - 1}{e^{x} (x^{2} + x)}$ $Now we reduces linear form,$ $Put, \frac{dv}{dx} = t \Rightarrow \frac{d^{2} v}{{dx}^{2}} = \frac{dt}{dx}$ $\frac{dt}{dx} + \frac{2 x + 1}{x^{2} + x} . t = \frac{x^{2} - x - 1}{e^{x} (x^{2} + x)}$ $IF = e^{\int Pdx} = e^{\int \frac{2 x + 1}{x^{2} + x} dx} = e^{\log (x^{2} + x)}$ $IF = x^{2} + x$ $t . IF = \int IF . Qdx + C_{1}$ $t . (x^{2} + x) = \int (x^{2} + x) . \frac{x^{2} - x - 1}{e^{x} (x^{2} + x)} dx + C_{1}$ $t (x^{2} + x) = \int \frac{x^{2} - x - 1}{e^{x}} dx + C_{1}$ $t (x^{2} + x) = \int x^{2} e^{- x} dx - \int {xe}^{- x} dx - \int e^{- x} dx + C_{1}$ $t (x^{2} + x) = [x^{2} . (- e^{- x}) - \int 2 x . (- e^{- x}) dx] - \int {xe}^{- x} dx - (- e^{- x}) + C_{1}$ $t (x^{2} + x) = - x^{2} e^{- x} + 2 \int {xe}^{- x} dx - \int {xe}^{- x} dx + e^{- x} + C_{1}$ $t (x^{2} + x) = - x^{2} e^{- x} + \int {xe}^{- x} dx + e^{- x} + C_{1}$ $t (x^{2} + x) = - x^{2} e^{- x} + x (- e^{x}) - \int 1 . (- e^{- x}) dx + e^{- x} + C_{1}$ $t (x^{2} + x) = - x^{2} e^{- x} - {xe}^{- x} - e^{- x} + e^{- x} + C_{1}$ $t (x^{2} + x) = - x^{2} e^{- x} - {xe}^{- x} + C_{1}$ $t = \frac{- {xe}^{- x} (x + 1)}{x (x + 1)} + C_{1} \frac{1}{x (x + 1)}$ $\frac{dv}{dx} = - e^{- x} + C_{1} \frac{1}{x (x + 1)}$ $\int dv = - \int e^{- x} dx + C_{1} \int \frac{1}{x (x + 1)} dx + C_{2}$ $v = - (- e^{- x}) + C_{1} [\int \frac{1}{x} dx - \int \frac{1}{x + 1} dx] + C_{2}$ $v = e^{- x} + C_{1} {\log x - \log (x + 1)} + C_{2}$ $v = e^{- x} + C_{1} \log \frac{x}{x + 1} + C_{2}$ $Now, again put value of v in {eq}^{n} (i)$ $y = e^{x} v . . . . . (i)$ $y = e^{x} (e^{- x} + C_{1} \log \frac{x}{x + 1} + C_{2})$ $y = 1 + C_{1} e^{x} \log \frac{x}{x + 1} + C_{2} e^{x} / / .$
	Commented by niroj last updated on 11/Jun/20
	thanks dear��
	Commented by bemath last updated on 11/Jun/20

	$waw === beautifull solution$
	Commented by bemath last updated on 11/Jun/20

	$thank you$
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