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Question Number 137069 by 0731619177 last updated on 29/Mar/21

	Answered by Ñï= last updated on 29/Mar/21
	$(1)y_p =(1/(D^2 +2D+5))(6sin 2x+7cos 2x) =(1/(2D+1))(6sin 2x+7cos 2x)=(((1+2D))/9)(6sin 2x+7cos 2x) =(1/9)(31cos 2x−22sin 2x) y=e^(−x) (C_1 sin 2x+C_2 cos 2x)+(1/9)(31cos 2x−22sin 2x) (3)y_p =(1/(D^2 +2D+1))e^x cos x=e^x (1/(4D+3))cos x=e^x ((3−4D)/(25))cos x =e^x ((3/(25))cos x+(4/(25))sin x) y=(C_1 +xC_2 )e^x +e^x ((3/(25))cos x+(4/(25))sin x) (9)y_p =(1/(D^2 +2D+2))(4e^(−x) x^2 sin x+3e^(−x) cos x) =e^(−x) (1/(D^2 +1))(4x^2 sin x+3cos x) =(2/i)e^(−x) (1/(D^2 +1))x^2 (e^(ix) −e^(−ix) )+xe^(−x) (3/((D^2 +1)′))cos x =(2/i)e^(−x) {e^(ix) (1/((D+i)^2 +1))x^2 −e^(−ix) (1/((D−i)^2 +1))x^2 }+(3/2)xe^(−x) sin x =(2/i)e^(−x) {e^(ix) (1/(D^2 +2iD))x^2 −e^(−ix) (1/(D^2 −2iD))x^2 }+(3/2)xe^(−x) sin x =(2/(3i))e^(−x) {e^(ix) (1/(D+2i))x^3 −e^(−ix) (1/(D−2i))x^3 }+(3/2)xe^(−x) sin x =−(1/3)e^(−x) {e^(ix) (1−(D/(2i))−(D^2 /4)+(D^3 /(8i)))x^3 +e^(−ix) (1+(D/(2i))−(D^2 /4)−(D^3 /(8i)))x^3 }+(3/2)xe^(−x) sin x =−(1/3)e^(−x) {e^(ix) (x^3 −(3/(2i))x^2 −(3/2)x+(3/(4i)))+e^(−ix) (x^3 +(3/(2i))x^2 −(3/2)x−(3/(4i)))}+(3/2)xe^(−x) sin x =−(1/3)e^(−x) (2x^3 cos x−3x^2 sin x−3xcos x+(3/2)sin x)+(3/2)xe^(−x) sin x y=e^(−x) (C_1 sin x+C_2 cos x)−(1/3)e^(−x) (2x^3 cos x−3x^2 sin x−3xcos x)+(3/2)xe^(−x) sin x$
	$(1) y_{p} = \frac{1}{D^{2} + 2 D + 5} (6 \sin 2 x + 7 \cos 2 x)$ $= \frac{1}{2 D + 1} (6 \sin 2 x + 7 \cos 2 x) = \frac{(1 + 2 D)}{9} (6 \sin 2 x + 7 \cos 2 x)$ $= \frac{1}{9} (31 \cos 2 x - 22 \sin 2 x)$ $y = e^{- x} (C_{1} \sin 2 x + C_{2} \cos 2 x) + \frac{1}{9} (31 \cos 2 x - 22 \sin 2 x)$ $(3) y_{p} = \frac{1}{D^{2} + 2 D + 1} e^{x} \cos x = e^{x} \frac{1}{4 D + 3} \cos x = e^{x} \frac{3 - 4 D}{25} \cos x$ $= e^{x} (\frac{3}{25} \cos x + \frac{4}{25} \sin x)$ $y = (C_{1} + {x C}_{2}) e^{x} + e^{x} (\frac{3}{25} \cos x + \frac{4}{25} \sin x)$ $(9) y_{p} = \frac{1}{D^{2} + 2 D + 2} (4 e^{- x} x^{2} \sin x + 3 e^{- x} \cos x)$ $= e^{- x} \frac{1}{D^{2} + 1} (4 x^{2} \sin x + 3 \cos x)$ $= \frac{2}{i} e^{- x} \frac{1}{D^{2} + 1} x^{2} (e^{i x} - e^{- i x}) + {x e}^{- x} \frac{3}{{(D^{2} + 1)}^{'}} \cos x$ $= \frac{2}{i} e^{- x} {e^{i x} \frac{1}{{(D + i)}^{2} + 1} x^{2} - e^{- i x} \frac{1}{{(D - i)}^{2} + 1} x^{2}} + \frac{3}{2} {x e}^{- x} \sin x$ $= \frac{2}{i} e^{- x} {e^{i x} \frac{1}{D^{2} + 2 i D} x^{2} - e^{- i x} \frac{1}{D^{2} - 2 i D} x^{2}} + \frac{3}{2} {x e}^{- x} \sin x$ $= \frac{2}{3 i} e^{- x} {e^{i x} \frac{1}{D + 2 i} x^{3} - e^{- i x} \frac{1}{D - 2 i} x^{3}} + \frac{3}{2} {x e}^{- x} \sin x$ $= - \frac{1}{3} e^{- x} {e^{i x} (1 - \frac{D}{2 i} - \frac{D^{2}}{4} + \frac{D^{3}}{8 i}) x^{3} + e^{- i x} (1 + \frac{D}{2 i} - \frac{D^{2}}{4} - \frac{D^{3}}{8 i}) x^{3}} + \frac{3}{2} {x e}^{- x} \sin x$ $= - \frac{1}{3} e^{- x} {e^{i x} (x^{3} - \frac{3}{2 i} x^{2} - \frac{3}{2} x + \frac{3}{4 i}) + e^{- i x} (x^{3} + \frac{3}{2 i} x^{2} - \frac{3}{2} x - \frac{3}{4 i})} + \frac{3}{2} {x e}^{- x} \sin x$ $= - \frac{1}{3} e^{- x} (2 x^{3} \cos x - 3 x^{2} \sin x - 3 x \cos x + \frac{3}{2} \sin x) + \frac{3}{2} {x e}^{- x} \sin x$ $y = e^{- x} (C_{1} \sin x + C_{2} \cos x) - \frac{1}{3} e^{- x} (2 x^{3} \cos x - 3 x^{2} \sin x - 3 x \cos x) + \frac{3}{2} {x e}^{- x} \sin x$
	Answered by Ñï= last updated on 29/Mar/21
	$(2)y_p =(1/(2D^2 +3D+1))(x^2 +3sin x) ={1−2D^2 −3D−(2D^2 +3D)^2 ...}x^2 +(1/(3D−1))3sin x =(1−3D−11D^2 −...)x^2 −(3/(10))(3D+1)sin x =x^2 −6x−22−(3/(10))(3cos x+sin x) y=C_2 e^(−x) +C_2 e^(−x/2) +x^2 −6x−22−(3/(10))(3cos x+sin x) (4)y_p =((12)/(D^2 +1))cos^2 x=(6/(D^2 +1))(cos 2x+1) =−(3/2)cos 2x+6 y=C_1 sin x+C_2 cos x−(3/2)cos 2x+6 (5)y_p =(2/(D^2 −3D+2))(x^2 +xe^x )=(1/(1+((D^2 −3D)/2)))x^2 +(2/(D^2 −3D+2))xe^x ={1−((D^2 −3D)/2)−(((D^2 −3D)^2 )/4)−...}x^2 +2e^x (1/(D(D−1)))x =(1+(3/2)D−((17)/(16))D^2 −...)x^2 +e^x (1/(D−1))x^2 =x^2 +3x−((17)/8)+e^x (1+D+D^2 +...)x^2 =x^2 +3x−((17)/8)+e^x (x^2 +2x+2) y=C_1 e^x +C_2 e^(2x) +e^x (x^2 +2x+2)$
	$(2) y_{p} = \frac{1}{2 D^{2} + 3 D + 1} (x^{2} + 3 \sin x)$ $= {1 - 2 D^{2} - 3 D - {(2 D^{2} + 3 D)}^{2} . . .} x^{2} + \frac{1}{3 D - 1} 3 \sin x$ $= (1 - 3 D - 11 D^{2} - . . .) x^{2} - \frac{3}{10} (3 D + 1) \sin x$ $= x^{2} - 6 x - 22 - \frac{3}{10} (3 \cos x + \sin x)$ $y = C_{2} e^{- x} + C_{2} e^{- x / 2} + x^{2} - 6 x - 22 - \frac{3}{10} (3 \cos x + \sin x)$ $(4) y_{p} = \frac{12}{D^{2} + 1} \cos^{2} x = \frac{6}{D^{2} + 1} (\cos 2 x + 1)$ $= - \frac{3}{2} \cos 2 x + 6$ $y = C_{1} \sin x + C_{2} \cos x - \frac{3}{2} \cos 2 x + 6$ $(5) y_{p} = \frac{2}{D^{2} - 3 D + 2} (x^{2} + {x e}^{x}) = \frac{1}{1 + \frac{D^{2} - 3 D}{2}} x^{2} + \frac{2}{D^{2} - 3 D + 2} {x e}^{x}$ $= {1 - \frac{D^{2} - 3 D}{2} - \frac{{(D^{2} - 3 D)}^{2}}{4} - . . .} x^{2} + 2 e^{x} \frac{1}{D (D - 1)} x$ $= (1 + \frac{3}{2} D - \frac{17}{16} D^{2} - . . .) x^{2} + e^{x} \frac{1}{D - 1} x^{2}$ $= x^{2} + 3 x - \frac{17}{8} + e^{x} (1 + D + D^{2} + . . .) x^{2}$ $= x^{2} + 3 x - \frac{17}{8} + e^{x} (x^{2} + 2 x + 2)$ $y = C_{1} e^{x} + C_{2} e^{2 x} + e^{x} (x^{2} + 2 x + 2)$
	Answered by Ñï= last updated on 29/Mar/21
	$(6)y_p =(1/(D^3 +D^2 +3D−5))(5sin 2x+10x^2 +3x+7) =−(1/(D+9))5sin 2x−(1/5)∙(1/(1−(1/5)(D^3 +D^2 +3D)))(10x^2 +3x+7) =(1/(17))(D−9)sin 2x−(1/5){1+(1/5)(D^3 +D^2 +3D)+(1/(25))(D^3 +D^2 +3D)^2 +...}(10x^2 +3x+7) =(1/(17))(2cos 2x−9sin 2x)−(1/5){1+(3/5)D+((14)/(25))D^2 +...}(10x^2 +3x+7) =(1/(17))(2cos 2x−9sin 2x)+2x^2 +3x+4 y=C_1 e^x +e^(−2x) (C_2 sin 2x+C_3 cos 2x)+(1/(17))(2cos 2x−9sin 2x)+2x^2 +3x+4 (7)y_p =(1/(D^3 +D))(2x^2 +4sin x)=(2/3)(1/(D^2 +1))x^3 −(1/(D^2 +1))4cos x =(2/3)(1−D^2 −D^4 −...)x^3 −x(1/((D^2 +1)′))4cos x =(2/3)(x^3 −6x)−2x(1/D)cos x =(2/3)x^3 −4x−2xsin x y=C_1 +C_2 sin x+C_3 cos x−(2/3)x^3 −4x−2xsin x$
	$(6) y_{p} = \frac{1}{D^{3} + D^{2} + 3 D - 5} (5 \sin 2 x + 10 x^{2} + 3 x + 7)$ $= - \frac{1}{D + 9} 5 \sin 2 x - \frac{1}{5} \cdot \frac{1}{1 - \frac{1}{5} (D^{3} + D^{2} + 3 D)} (10 x^{2} + 3 x + 7)$ $= \frac{1}{17} (D - 9) \sin 2 x - \frac{1}{5} {1 + \frac{1}{5} (D^{3} + D^{2} + 3 D) + \frac{1}{25} {(D^{3} + D^{2} + 3 D)}^{2} + . . .} (10 x^{2} + 3 x + 7)$ $= \frac{1}{17} (2 \cos 2 x - 9 \sin 2 x) - \frac{1}{5} {1 + \frac{3}{5} D + \frac{14}{25} D^{2} + . . .} (10 x^{2} + 3 x + 7)$ $= \frac{1}{17} (2 \cos 2 x - 9 \sin 2 x) + 2 x^{2} + 3 x + 4$ $y = C_{1} e^{x} + e^{- 2 x} (C_{2} \sin 2 x + C_{3} \cos 2 x) + \frac{1}{17} (2 \cos 2 x - 9 \sin 2 x) + 2 x^{2} + 3 x + 4$ $(7) y_{p} = \frac{1}{D^{3} + D} (2 x^{2} + 4 \sin x) = \frac{2}{3} \frac{1}{D^{2} + 1} x^{3} - \frac{1}{D^{2} + 1} 4 \cos x$ $= \frac{2}{3} (1 - D^{2} - D^{4} - . . .) x^{3} - x \frac{1}{{(D^{2} + 1)}^{'}} 4 \cos x$ $= \frac{2}{3} (x^{3} - 6 x) - 2 x \frac{1}{D} \cos x$ $= \frac{2}{3} x^{3} - 4 x - 2 x \sin x$ $y = C_{1} + C_{2} \sin x + C_{3} \cos x - \frac{2}{3} x^{3} - 4 x - 2 x \sin x$
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