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D-3-2D-2-9D-18-y-6cos3x-




Question Number 56580 by subhankar10 last updated on 18/Mar/19
(D^3 −2D^2 +9D−18)y=6cos3x
(D32D2+9D18)y=6cos3x
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Mar/19
let y=e^(mx)  be a solution  Dy=me^(mx)   D^2 y=m^2 e^(mx)   ..  so  complimentary function    m^3 e^(mx) −2m^2 e^(mx) +9me^(mx) −18e^(mx) =0  e^(mx) (m^3 −2m^2 +9m−18)=0  e^(mx) ≠0  m^3 −2m^2 +9m−18=0  m^2 (n−2)+9(m−2)=0  (m−2)(m^2 +9)=0  m=2 and  ±3i  C.F  C_1 e^(2x) +C_2 e^(i3x) +C_3 e^(−i3x)   Particular intregal  y=((6cos3x)/(D^3 −2D^2 +9D−18))  =6×((cos3x)/(D^2 (D−2)+9(D−2)))  =6×((cos3x)/((D−2)(D^2 +9)))  =6×(((D+2) cos3x)/((D^2 −4)(D^2 +9)))  =(6/((−3^2 −4)))×((−3sin3x+2cos3x)/(D^2 +9))  =((−6)/(13))×((−3sin3x+2cos3x)/(D^2 +9))  =(6/(13))×(((3sin3x−2cos3x))/(D^2 +9))  =(6/(13))×((rsin(3x−θ))/(D^2 +9)) [r=(√(3^2 +2^2 )) =(√(13))   tanθ=(2/3)]  p=(6/( (√(13)) ))×((cos(3x−θ))/(D^2 +9))  q=(6/( (√(13))))×((sin(3x−θ))/(D^2 +9))  p+iq=(6/( (√(13))))×(e^(i(3x−θ)) /(D^2 +9))  p+iq=(6/( (√(13))))×e^(−iθ) ×(e^(i×3x) /(D^2 +9))  p+iq=(6/( (√(13))))×e^(−iθ) ×(e^(i3x) /((D+i3)(D−i3)))  p+iq=(6/( (√(13))))×(cosθ−isinθ)×(e^(i3x) /((i3+i3)(D+i3−i3)))×1  =(6/( (√(13))))×((3/( (√(13))))−i×(2/( (√(13)))))×(((cos3x+isin3x))/(6i))×x  =(1/(13))×((3i+2)/i^2 )×(xcos3x+ixsin3x)  =((−1)/(13))(i3xcos3x−3xsin3x+2xcos3x+i2xsin3x)  =((−1)/(13)){(2xcos3x−3xsin3x)+i(3xcos3x+2xsin3x)}  =((−x)/(13))(2cos3x−3sin3x)+((−ix)/(13))(3cos3x+2sin3x)  now our answer is related to complex part  so answer is  Particular intregal is  =((−x)/(13))(3cos3x+2sin3x)  so complte answer is=C.F+P.I  y=C_1 e^(2x) +C_2 e^(i3x) +C_3 e^(−i3x) +((−x)/(13))(3cos3x+2sin3x)
lety=emxbeasolutionDy=memxD2y=m2emx..socomplimentaryfunctionm3emx2m2emx+9memx18emx=0emx(m32m2+9m18)=0emx0m32m2+9m18=0m2(n2)+9(m2)=0(m2)(m2+9)=0m=2and±3iC.FC1e2x+C2ei3x+C3ei3xParticularintregaly=6cos3xD32D2+9D18=6×cos3xD2(D2)+9(D2)=6×cos3x(D2)(D2+9)=6×(D+2)cos3x(D24)(D2+9)=6(324)×3sin3x+2cos3xD2+9=613×3sin3x+2cos3xD2+9=613×(3sin3x2cos3x)D2+9=613×rsin(3xθ)D2+9[r=32+22=13tanθ=23]p=613×cos(3xθ)D2+9q=613×sin(3xθ)D2+9p+iq=613×ei(3xθ)D2+9p+iq=613×eiθ×ei×3xD2+9p+iq=613×eiθ×ei3x(D+i3)(Di3)p+iq=613×(cosθisinθ)×ei3x(i3+i3)(D+i3i3)×1=613×(313i×213)×(cos3x+isin3x)6i×x=113×3i+2i2×(xcos3x+ixsin3x)=113(i3xcos3x3xsin3x+2xcos3x+i2xsin3x)=113{(2xcos3x3xsin3x)+i(3xcos3x+2xsin3x)}=x13(2cos3x3sin3x)+ix13(3cos3x+2sin3x)nowouranswerisrelatedtocomplexpartsoanswerisParticularintregalis=x13(3cos3x+2sin3x)socomplteansweris=C.F+P.Iy=C1e2x+C2ei3x+C3ei3x+x13(3cos3x+2sin3x)

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