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Question-135173

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Question Number 135173 by bemath last updated on 11/Mar/21
Answered by john_santu last updated on 11/Mar/21
Homogenous problem y_h ′′′−y_h ′′−2y_h ′ = 0 let y_h = e^(mx) to get m^3 −m^2 −2m = 0 m(m^2 −m−2) = 0 m=0 , m=−1 , m=2 Homogenous solution y_h = C_1 +C_2 e^(−x) +C_3 e^(2x) Particular solution y_p = asin (2x)+bcos (2x)+cxe^(−x) solving for coefficients we get y_p = −(3/(10))sin (2x)+(1/(10))cos (2x)+(1/3)xe^(−x) The General solution y=C_1 +C_2 e^(−x) +C_3 e^(2x) −{((3sin (2x)−cos (2x))/(10))}+((xe^(−x) )/3) computing y(0)=y′(0)=0 , y′′(0)=1 we get C_1 =−1, C_2 =((23)/(45)) and C_3 =(7/(18)) Therefore solution is y = −1+((23)/(45))e^(−x) +(7/(18))e^(2x) −{((3sin (2x)−cos (2x))/(10))}+((xe^(−x) )/3)
$${Homogenous}\:{problem} \\ $$$${y}_{{h}} '''−{y}_{{h}} ''−\mathrm{2}{y}_{{h}} '\:=\:\mathrm{0} \\ $$$${let}\:{y}_{{h}} \:=\:{e}^{{mx}} \:{to}\:{get}\:{m}^{\mathrm{3}} −{m}^{\mathrm{2}} −\mathrm{2}{m}\:=\:\mathrm{0} \\ $$$${m}\left({m}^{\mathrm{2}} −{m}−\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$${m}=\mathrm{0}\:,\:{m}=−\mathrm{1}\:,\:{m}=\mathrm{2} \\ $$$${Homogenous}\:{solution}\: \\ $$$${y}_{{h}} \:=\:{C}_{\mathrm{1}} +{C}_{\mathrm{2}} {e}^{−{x}} +{C}_{\mathrm{3}} {e}^{\mathrm{2}{x}} \\ $$$${Particular}\:{solution} \\ $$$${y}_{{p}} =\:{a}\mathrm{sin}\:\left(\mathrm{2}{x}\right)+{b}\mathrm{cos}\:\left(\mathrm{2}{x}\right)+{cxe}^{−{x}} \\ $$$${solving}\:{for}\:{coefficients} \\ $$$${we}\:{get}\:{y}_{{p}} =\:−\frac{\mathrm{3}}{\mathrm{10}}\mathrm{sin}\:\left(\mathrm{2}{x}\right)+\frac{\mathrm{1}}{\mathrm{10}}\mathrm{cos}\:\left(\mathrm{2}{x}\right)+\frac{\mathrm{1}}{\mathrm{3}}{xe}^{−{x}} \\ $$$${The}\:\mathcal{G}{eneral}\:{solution} \\ $$$${y}={C}_{\mathrm{1}} +{C}_{\mathrm{2}} {e}^{−{x}} +{C}_{\mathrm{3}} {e}^{\mathrm{2}{x}} −\left\{\frac{\mathrm{3sin}\:\left(\mathrm{2}{x}\right)−\mathrm{cos}\:\left(\mathrm{2}{x}\right)}{\mathrm{10}}\right\}+\frac{{xe}^{−{x}} }{\mathrm{3}} \\ $$$${computing}\:{y}\left(\mathrm{0}\right)={y}'\left(\mathrm{0}\right)=\mathrm{0}\:,\:{y}''\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${we}\:{get}\:{C}_{\mathrm{1}} =−\mathrm{1},\:{C}_{\mathrm{2}} =\frac{\mathrm{23}}{\mathrm{45}}\:{and}\:{C}_{\mathrm{3}} =\frac{\mathrm{7}}{\mathrm{18}} \\ $$$${Therefore}\:{solution}\:{is} \\ $$$${y}\:=\:−\mathrm{1}+\frac{\mathrm{23}}{\mathrm{45}}{e}^{−{x}} +\frac{\mathrm{7}}{\mathrm{18}}{e}^{\mathrm{2}{x}} −\left\{\frac{\mathrm{3sin}\:\left(\mathrm{2}{x}\right)−\mathrm{cos}\:\left(\mathrm{2}{x}\right)}{\mathrm{10}}\right\}+\frac{{xe}^{−{x}} }{\mathrm{3}} \\ $$$$ \\ $$
Answered by Ñï= last updated on 11/Mar/21
y_p =(1/(D^3 −D^2 −2D))e^(−x) +(1/(D^3 −D^2 −2D))4cos 2x =x(1/((D^3 −D^2 −2D)′))e^(−x) +2(1/(D^2 −D−2))sin 2x =x(1/(3D^2 −2D−2))e^(−x) +2((D^2 −2+D)/((D^2 −2)^2 −D^2 ))sin 2x =(1/3)xe^(−x) +(((−6+D))/(20))sin 2x =(1/3)xe^(−x) +(1/(10))cos 2x−(3/(10))sin 2x y=C_1 +C_2 e^(−x) +C_3 e^(2x) +(1/3)xe^(−x) +(1/(10))cos 2x−(3/(10))sin 2x
$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{3}} −{D}^{\mathrm{2}} −\mathrm{2}{D}}{e}^{−{x}} +\frac{\mathrm{1}}{{D}^{\mathrm{3}} −{D}^{\mathrm{2}} −\mathrm{2}{D}}\mathrm{4cos}\:\mathrm{2}{x} \\ $$$$={x}\frac{\mathrm{1}}{\left({D}^{\mathrm{3}} −{D}^{\mathrm{2}} −\mathrm{2}{D}\right)'}{e}^{−{x}} +\mathrm{2}\frac{\mathrm{1}}{{D}^{\mathrm{2}} −{D}−\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x} \\ $$$$={x}\frac{\mathrm{1}}{\mathrm{3}{D}^{\mathrm{2}} −\mathrm{2}{D}−\mathrm{2}}{e}^{−{x}} +\mathrm{2}\frac{{D}^{\mathrm{2}} −\mathrm{2}+{D}}{\left({D}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} −{D}^{\mathrm{2}} }\mathrm{sin}\:\mathrm{2}{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{xe}^{−{x}} +\frac{\left(−\mathrm{6}+{D}\right)}{\mathrm{20}}\mathrm{sin}\:\mathrm{2}{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{xe}^{−{x}} +\frac{\mathrm{1}}{\mathrm{10}}{cos}\:\mathrm{2}{x}−\frac{\mathrm{3}}{\mathrm{10}}\mathrm{sin}\:\mathrm{2}{x} \\ $$$${y}={C}_{\mathrm{1}} +{C}_{\mathrm{2}} {e}^{−{x}} +{C}_{\mathrm{3}} {e}^{\mathrm{2}{x}} +\frac{\mathrm{1}}{\mathrm{3}}{xe}^{−{x}} +\frac{\mathrm{1}}{\mathrm{10}}\mathrm{cos}\:\mathrm{2}{x}−\frac{\mathrm{3}}{\mathrm{10}}\mathrm{sin}\:\mathrm{2}{x} \\ $$

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