solve y^(′′) −2y^′ +y =xe^(−x) sin(2x) withy(o) =−1 and y^′ (0) =0


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Question Number 99464 by mathmax by abdo last updated on 21/Jun/20

$solve y^{^{″}} - {2 y}^{^{'}} + y = {xe}^{- x} \sin (2 x) withy (o) = - 1 and y^{^{'}} (0) = 0$
	Answered by MWSuSon last updated on 21/Jun/20

	$i do not know how to use D - operator$ $method to find the particular integral .$ $but if i make . . .$ $y_{p} = {Ae}^{- x} xsin (2 x) + {Be}^{- x} \sin (2 x) + C$ $e^{- x} xcos (2 x) + {De}^{- x} \cos (2 x)$ $and equate coefficients of y_{p}^{'}, y_{p}^{″} and y_{p} .$ $This will take more time . I am just$ $dropping this comment so that i can have$ $notification when someone solves it using$ $D - operator method .$
	Answered by mathmax by abdo last updated on 22/Jun/20
	$let solve by laplace e ⇒L(y^(′′) )−2L(y^′ )+L(y) = L(xe^(−x) sin(2x)) ⇒ x^2 L(y)−xy(0)−y^′ (0)−2(xL(y)−y(o))+L(y) =L(xe^(−x) sin(2x)) ⇒ (x^2 −2x +1)L(y) +x−2 =L(xe^(−x) sin(2x)) we have L(xe^(−x) sin(2x)) =∫_0 ^∞ t e^(−t) sin(2t)e^(−xt) dt =Im(∫_0 ^∞ t e^(−t+2it−xt) dt) =Im(∫_0 ^∞ t e^((−x−1+2i)t) dt) and ∫_0 ^∞ t e^((−x−1+2i)t) dt =_(byparts) [(t/(−x−1+2i))e^((−x−1+2i)t) ]_0 ^∞ −∫_0 ^∞ (1/(−x−1+2i))e^((−x−1+2i)t) dt =(1/(x+1−2i))∫_0 ^∞ e^((−x−1+2i)t) dt =(1/((x+1−2i)))[(1/(−x−1+2i))e^((−x−1+2i)t) ]_0 ^∞ =(1/((x+1−2i)^2 )) =(((x+1+2i)^2 )/({(x+1)^2 +4}^2 )) =(((x+1)^2 +4i(x+1)−4)/({(x+1)^2 +4}^2 )) ⇒ Im(....) =((4(x+1))/({(x+1)^2 +4}^2 )) e⇒(x^2 −2x+1)L(y) =−x+2 +((4x+4)/({(x+1)^2 +4}^2 )) ⇒ L(y) =((−x+2)/(x^2 −2x+1)) +((4x+4)/((x^2 −2x+1){(x+1)^2 +4}^2 )) ⇒ L(y) =((−x+2)/((x−1)^2 )) +((4x+4)/((x−1)^2 {(x+1)^2 +4}^2 )) ⇒ y(x) =L^(−1) (((−x+2)/((x−1)^2 ))) +4 L^(−1) (((x+1)/((x−1)^2 {(x+1)^2 +4}))) ...be continued....$
	$let solve by laplace e \Rightarrow L (y^{^{″}}) - 2 L (y^{^{'}}) + L (y) = L ({xe}^{- x} \sin (2 x)) \Rightarrow$ $x^{2} L (y) - xy (0) - y^{^{'}} (0) - 2 (xL (y) - y (o)) + L (y) = L ({xe}^{- x} \sin (2 x)) \Rightarrow$ $(x^{2} - 2 x + 1) L (y) + x - 2 = L ({xe}^{- x} \sin (2 x))$ $we have L ({xe}^{- x} \sin (2 x)) = \int_{0}^{\infty} t e^{- t} \sin (2 t) e^{- xt} dt$ $= Im (\int_{0}^{\infty} t e^{- t + 2 it - xt} dt) = Im (\int_{0}^{\infty} t e^{(- x - 1 + 2 i) t} dt) and$ $\int_{0}^{\infty} t e^{(- x - 1 + 2 i) t} dt =_{byparts} {[\frac{t}{- x - 1 + 2 i} e^{(- x - 1 + 2 i) t}]}_{0}^{\infty} - \int_{0}^{\infty} \frac{1}{- x - 1 + 2 i} e^{(- x - 1 + 2 i) t} dt$ $= \frac{1}{x + 1 - 2 i} \int_{0}^{\infty} e^{(- x - 1 + 2 i) t} dt = \frac{1}{(x + 1 - 2 i)} {[\frac{1}{- x - 1 + 2 i} e^{(- x - 1 + 2 i) t}]}_{0}^{\infty}$ $= \frac{1}{{(x + 1 - 2 i)}^{2}} = \frac{{(x + 1 + 2 i)}^{2}}{{{(x + 1)}^{2} + 4}^{2}} = \frac{{(x + 1)}^{2} + 4 i (x + 1) - 4}{{{(x + 1)}^{2} + 4}^{2}} \Rightarrow$ $Im (. . . .) = \frac{4 (x + 1)}{{{(x + 1)}^{2} + 4}^{2}}$ $e \Rightarrow (x^{2} - 2 x + 1) L (y) = - x + 2 + \frac{4 x + 4}{{{(x + 1)}^{2} + 4}^{2}} \Rightarrow$ $L (y) = \frac{- x + 2}{x^{2} - 2 x + 1} + \frac{4 x + 4}{(x^{2} - 2 x + 1) {{(x + 1)}^{2} + 4}^{2}} \Rightarrow$ $L (y) = \frac{- x + 2}{{(x - 1)}^{2}} + \frac{4 x + 4}{{(x - 1)}^{2} {{(x + 1)}^{2} + 4}^{2}} \Rightarrow$ $y (x) = L^{- 1} (\frac{- x + 2}{{(x - 1)}^{2}}) + 4 L^{- 1} (\frac{x + 1}{{(x - 1)}^{2} {{(x + 1)}^{2} + 4}})$ $. . . be continued . . . .$
	Commented by mathmax by abdo last updated on 22/Jun/20
	$y(x) =L^(−1) (((−x+2)/((x−1)^2 ))) +4 L^(−1) (((x+1)/((x−1)^2 {(x+1)^2 +4}^2 )))$
	$y (x) = L^{- 1} (\frac{- x + 2}{{(x - 1)}^{2}}) + 4 L^{- 1} (\frac{x + 1}{{(x - 1)}^{2} {{(x + 1)}^{2} + 4}^{2}})$
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