solve (D^2 −2D+1)y = e^x ln x by using the method of variation of parameters.


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Question Number 119065 by benjo_mathlover last updated on 21/Oct/20

$s o l v e (D^{2} - 2 D + 1) y = e^{x} \ln x$ $b y u s i n g t h e m e t h o d o f v a r i a t i o n$ $o f p a r a m e t e r s .$
	Answered by Olaf last updated on 22/Oct/20

	$y = e^{x} u$ $y^{'} = (u^{'} + u) e^{x}$ $y^{″} = (u^{″} + 2 u^{'} + u) e^{x}$ $u^{″} + 2 u^{'} + u - 2 (u^{'} + u) + u = \ln x$ $u^{″} = \ln x$ $u^{'} = \int \ln x d x = x \ln x - x + C_{1}$ $u = \int (x \ln x - x + C_{1}) d x$ $u = \frac{x^{2}}{2} \ln x - \int \frac{x^{2}}{2} . \frac{1}{x} d x - \frac{x^{2}}{2} + C_{1} x + C_{2}$ $u = \frac{x^{2}}{2} \ln x - \frac{x^{2}}{4} - \frac{x^{2}}{2} + C_{1} x + C_{2}$ $u = \frac{x^{2}}{2} \ln x - \frac{3 x^{2}}{4} + C_{1} x + C_{2}$ $y = (\frac{x^{2}}{2} \ln x - \frac{3 x^{2}}{4} + C_{1} x + C_{2}) e^{x}$
	Commented by benjo_mathlover last updated on 22/Oct/20

	$t h a n k y o u$
	Answered by john santu last updated on 27/Oct/20
	$For the homogenous solution we solve the characteristic equation λ^2 −2λ+1 = 0 , which has a double root λ=1 hence the homogenous solution can be written as y_h = (C_1 +C_2 x)e^x using the homogenous solution we can now begin using variation of parameters method let y_p = u.e^x + v.xe^x Differentiating yields y ′ = (ue^x +u′e^x )+(v.(x+1)e^x +v′.xe^x ) set u′e^x + v′xe^x = 0 or u′+xv′ = 0 we get y′ = ue^x +v.(x+1)e^x Differentiating again gives y′′ = ue^x +u′e^x +v(x+2)e^x +v′(x+1)e^x substituting into the original differential equation (ue^x +u′e^x +v(x+2)e^x +v′(x+1)e^x )− 2(ue^x +v(x+1)e^x ) + (ue^x +v.xe^x ) = e^x .ln x this reduces to u′+(x+1)v′ = ln x now we have two equation in u′ and v′ { ((u′+xv′ = 0)),((u′+(x+1)v′ = ln x)) :} (((1 x)),((1 x+1)) ) (((u′)),((v′)) ) = ((( 0)),((ln x)) ) → { ((u′ = −x ln x)),((v′ = ln x)) :} integrating with by parts gives { ((u = −(1/2)x^2 ln x + (1/4)x^(2 ) )),((v = x ln x−x )) :} thus particular solution equal to y_p = −(1/2)x^2 e^x ln x + (1/4)x^2 e^x + x^2 e^x ln x −x^2 e^x y_p = (1/2)x^2 e^x ln x −(3/4)x^2 e^x Therefore we conclude that a general solution y = y_h + y_p y = (C_1 +C_2 x)e^x + (1/4)(2ln x−3)x^2 e^x$
	$F o r t h e h o m o g e n o u s s o l u t i o n w e s o l v e t h e$ $c h a r a c t e r i s t i c e q u a t i o n$ $λ^{2} - 2 λ + 1 = 0, w h i c h h a s a d o u b l e r o o t λ = 1$ $h e n c e t h e h o m o g e n o u s s o l u t i o n c a n b e$ $w r i t t e n a s y_{h} = (C_{1} + C_{2} x) e^{x}$ $u s i n g t h e h o m o g e n o u s s o l u t i o n w e$ $c a n n o w b e g i n u s i n g v a r i a t i o n o f$ $p a r a m e t e r s m e t h o d$ $l e t y_{p} = u . e^{x} + v . {x e}^{x}$ $D i f f e r e n t i a t i n g y i e l d s$ $y^{'} = ({u e}^{x} + u^{'} e^{x}) + (v . (x + 1) e^{x} + v^{'} . {x e}^{x})$ $s e t u^{'} e^{x} + v^{'} {x e}^{x} = 0 o r u^{'} + {x v}^{'} = 0$ $w e g e t y^{'} = {u e}^{x} + v . (x + 1) e^{x}$ $D i f f e r e n t i a t i n g a g a i n g i v e s$ $y^{″} = {u e}^{x} + u^{'} e^{x} + v (x + 2) e^{x} + v^{'} (x + 1) e^{x}$ $s u b s t i t u t i n g i n t o t h e o r i g i n a l$ $d i f f e r e n t i a l e q u a t i o n$ $({u e}^{x} + u^{'} e^{x} + v (x + 2) e^{x} + v^{'} (x + 1) e^{x}) -$ $2 ({u e}^{x} + v (x + 1) e^{x}) + ({u e}^{x} + v . {x e}^{x}) = e^{x} . \ln x$ $t h i s r e d u c e s t o$ $u^{'} + (x + 1) v^{'} = \ln x$ $n o w w e h a v e t w o e q u a t i o n i n u^{'} a n d v^{'}$ ${\begin{cases} u^{'} + {x v}^{'} = 0 \\ u^{'} + (x + 1) v^{'} = \ln x \end{cases}$ $(\begin{matrix} 1 x \\ 1 x + 1 \end{matrix}) (\begin{matrix} u^{'} \\ v^{'} \end{matrix}) = (\begin{matrix} 0 \\ \ln x \end{matrix})$ $\to {\begin{cases} u^{'} = - x \ln x \\ v^{'} = \ln x \end{cases}$ $i n t e g r a t i n g w i t h b y p a r t s g i v e s$ ${\begin{cases} u = - \frac{1}{2} x^{2} \ln x + \frac{1}{4} x^{2} \\ v = x \ln x - x \end{cases}$ $t h u s p a r t i c u l a r s o l u t i o n e q u a l t o$ $y_{p} = - \frac{1}{2} x^{2} e^{x} \ln x + \frac{1}{4} x^{2} e^{x}$ $+ x^{2} e^{x} \ln x - x^{2} e^{x}$ $y_{p} = \frac{1}{2} x^{2} e^{x} \ln x - \frac{3}{4} x^{2} e^{x}$ $T h e r e f o r e w e c o n c l u d e t h a t a g e n e r a l$ $s o l u t i o n y = y_{h} + y_{p}$ $y = (C_{1} + C_{2} x) e^{x} + \frac{1}{4} (2 \ln x - 3) x^{2} e^{x}$
	Answered by TANMAY PANACEA last updated on 22/Oct/20

	$P . I = \frac{e^{x} l n x}{{(D - 1)}^{2}} = e^{x} \times \frac{1}{{(D + 1 - 1)}^{2}} \times l n x$ $= \frac{e^{x}}{1} \times \frac{l n x}{D^{2}}$ $\frac{1}{D} \times l n x = x l n x - x$ $\frac{1}{D^{2}} l n x = \int (x l n x - x) d x$ $= l n x \times \frac{x^{2}}{2} - \int \frac{1}{x} \times \frac{x^{2}}{2} d x - \frac{x^{2}}{2}$ $= \frac{x^{2} l n x}{2} - \frac{x^{2}}{4} - \frac{x^{2}}{2}$ $= \frac{x^{2}}{2} (l n x - \frac{1}{2} - 1) = \frac{x^{2}}{2} (\frac{- 3}{2} + l n x)$ $s n s s e r$ $\frac{e^{x} \times x^{2}}{2} o (\frac{- 3}{2} + l n x)$
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