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Question Number 135402 by mohammad17 last updated on 12/Mar/21

	Answered by Ñï= last updated on 13/Mar/21
	$y_p =(1/(D^2 −2D+2))(4xe^x cos x+xe^(−x) +x^2 +1) 1>>(1/(D^2 −2D+2))4xe^x cos x=2(1/(D^2 −2D+2))e^x (e^(ix) +e^(−ix) )x =2{e^((1+i)x) (1/((D+1+i)^2 −2(D+1+i)+2))x+e^((1−i)x) (1/((D+1−i)^2 −2(D+1−i)+2))x} =2{e^((1+i)x) (1/((D+2i)D))x+e^((1−i)x) (1/((D−2i)D))x}={e^((1+i)x) (1/(D+2i))x^2 +e^((1−i)x) (1/(D−2i))x^2 } =e^((1+i)x) (1/(2i(1+(D/(2i)))))x^2 −e^((1−i)x) (1/(2i(1−(D/(2i)))))x^2 =(1/(2i))e^((1+i)x) {1+(−(D/(2i)))+(−(D/(2i)))^2 }x^2 −(e^((1−i)x) /(2i)){1+((D/(2i)))+((D/(2i)))^2 }x^2 =(1/(2i))e^((1+i)x) {x^2 −(x/i)−(1/2)}−(1/(2i))e^((1−i)x) {x^2 +(x/i)−(1/2)} =e^((1+i)x) {(x^2 /(2i))+(x/2)−(1/(4i))}+e^((1−i)x) {−(x^2 /(2i))+(x/2)+(1/(4i))} =e^x {e^(ix) (x^2 /(2i))+e^(ix) (x/2)−e^(ix) (1/(4i))−e^(−ix) (x^2 /(2i))+e^(−ix) (x/2)+e^(−ix) (1/(4i))} =e^x (xcos x−(1/2)sin x+x^2 sin x) 2>>(1/(D^2 −2D+2))xe^(−x) =e^(−x) (1/((D−1)^2 −2(D−1)+2))x =e^(−x) (1/(D^2 −4D+5))x=e^(−x) (1/(5(1+(D^2 /5)−((4D)/5))))x =(e^(−x) /5)(1+((4D)/5)−(D^2 /5))x=(e^(−x) /5)(x+(4/5)) 3>>(1/(D^2 −2D+2))(x^2 +1)=(1/(2(1−D+(D^2 /2))))(x^2 +1) =(1/2)(1+D−(D^2 /2)+(D−(D^2 /2))^2 +...)(x^2 +1) =(1/2)(1+D+(D^2 /2)+...)(x^2 +1)=(1/2)(x^2 +2x+2) y_p =e^x (xcos x−(1/2)sin x+x^2 sin x)+(1/5)(x+(4/5))e^(−x) +(1/2)(x^2 +2x) λ^2 −2λ+2=0 ⇒λ=1±i ⇒y_h =e^x (C_1 ′sin x+C_2 cos x) y=y_h +y_p =e^x (C_1 sin x+C_2 cos x+xcos x+x^2 sin x)+(1/5)(x+(4/5))e^(−x) +(1/2)(x^2 +2x+2)$
	$y_{p} = \frac{1}{D^{2} - 2 D + 2} (4 {x e}^{x} co s x + {x e}^{- x} + x^{2} + 1)$ $1 >> \frac{1}{D^{2} - 2 D + 2} 4 {x e}^{x} c o s x = 2 \frac{1}{D^{2} - 2 D + 2} e^{x} (e^{i x} + e^{- i x}) x$ $= 2 {e^{(1 + i) x} \frac{1}{{(D + 1 + i)}^{2} - 2 (D + 1 + i) + 2} x + e^{(1 - i) x} \frac{1}{{(D + 1 - i)}^{2} - 2 (D + 1 - i) + 2} x}$ $= 2 {e^{(1 + i) x} \frac{1}{(D + 2 i) D} x + e^{(1 - i) x} \frac{1}{(D - 2 i) D} x} = {e^{(1 + i) x} \frac{1}{D + 2 i} x^{2} + e^{(1 - i) x} \frac{1}{D - 2 i} x^{2}}$ $= e^{(1 + i) x} \frac{1}{2 i (1 + \frac{D}{2 i})} x^{2} - e^{(1 - i) x} \frac{1}{2 i (1 - \frac{D}{2 i})} x^{2}$ $= \frac{1}{2 i} e^{(1 + i) x} {1 + (- \frac{D}{2 i}) + {(- \frac{D}{2 i})}^{2}} x^{2} - \frac{e^{(1 - i) x}}{2 i} {1 + (\frac{D}{2 i}) + {(\frac{D}{2 i})}^{2}} x^{2}$ $= \frac{1}{2 i} e^{(1 + i) x} {x^{2} - \frac{x}{i} - \frac{1}{2}} - \frac{1}{2 i} e^{(1 - i) x} {x^{2} + \frac{x}{i} - \frac{1}{2}}$ $= e^{(1 + i) x} {\frac{x^{2}}{2 i} + \frac{x}{2} - \frac{1}{4 i}} + e^{(1 - i) x} {- \frac{x^{2}}{2 i} + \frac{x}{2} + \frac{1}{4 i}}$ $= e^{x} {e^{i x} \frac{x^{2}}{2 i} + e^{i x} \frac{x}{2} - e^{i x} \frac{1}{4 i} - e^{- i x} \frac{x^{2}}{2 i} + e^{- i x} \frac{x}{2} + e^{- i x} \frac{1}{4 i}}$ $= e^{x} (x \cos x - \frac{1}{2} s i n x + x^{2} s i n x)$ $2 >> \frac{1}{D^{2} - 2 D + 2} {x e}^{- x} = e^{- x} \frac{1}{{(D - 1)}^{2} - 2 (D - 1) + 2} x$ $= e^{- x} \frac{1}{D^{2} - 4 D + 5} x = e^{- x} \frac{1}{5 (1 + \frac{D^{2}}{5} - \frac{4 D}{5})} x$ $= \frac{e^{- x}}{5} (1 + \frac{4 D}{5} - \frac{D^{2}}{5}) x = \frac{e^{- x}}{5} (x + \frac{4}{5})$ $3 >> \frac{1}{D^{2} - 2 D + 2} (x^{2} + 1) = \frac{1}{2 (1 - D + \frac{D^{2}}{2})} (x^{2} + 1)$ $= \frac{1}{2} (1 + D - \frac{D^{2}}{2} + {(D - \frac{D^{2}}{2})}^{2} + . . .) (x^{2} + 1)$ $= \frac{1}{2} (1 + D + \frac{D^{2}}{2} + . . .) (x^{2} + 1) = \frac{1}{2} (x^{2} + 2 x + 2)$ $y_{p} = e^{x} (x \cos x - \frac{1}{2} \sin x + x^{2} s i n x) + \frac{1}{5} (x + \frac{4}{5}) e^{- x} + \frac{1}{2} (x^{2} + 2 x)$ $λ^{2} - 2 λ + 2 = 0$ $\Rightarrow λ = 1 \pm i$ $\Rightarrow y_{h} = e^{x} (C_{1}^{'} \sin x + C_{2} \cos x)$ $y = y_{h} + y_{p} = e^{x} (C_{1} \sin x + C_{2} \cos x + x \cos x + x^{2} \sin x) + \frac{1}{5} (x + \frac{4}{5}) e^{- x} + \frac{1}{2} (x^{2} + 2 x + 2)$
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