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Question Number 136030 by mathmax by abdo last updated on 18/Mar/21
solve y^((3)) −2y^((2))  +y =x−sinx
$$\mathrm{solve}\:\mathrm{y}^{\left(\mathrm{3}\right)} −\mathrm{2y}^{\left(\mathrm{2}\right)} \:+\mathrm{y}\:=\mathrm{x}−\mathrm{sinx} \\ $$
Answered by Ñï= last updated on 18/Mar/21
y′′′−2y′′+y=x−sin x  y_p =(1/(D^3 −2D^2 +1))(x−sin x)  =x−(1/(D^3 −2D^2 +1))sin x  =x−(1/(3−D))sin x  =x−(1/(10))(3+D)sin x  =x−(1/(10))(3sin x+cos x)  λ^3 −2λ^2 +1=0  ⇒λ_1 =1      λ_2 =((1+(√5))/2)       λ_3 =((1−(√5))/2)  ⇒y=C_1 e^x +C_2 e^(((1+(√5))/2)x) +C_3 e^(((1−(√5))/2)x) +x−(1/(10))(3sin x+cos x)
$${y}'''−\mathrm{2}{y}''+{y}={x}−\mathrm{sin}\:{x} \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{3}} −\mathrm{2}{D}^{\mathrm{2}} +\mathrm{1}}\left({x}−\mathrm{sin}\:{x}\right) \\ $$$$={x}−\frac{\mathrm{1}}{{D}^{\mathrm{3}} −\mathrm{2}{D}^{\mathrm{2}} +\mathrm{1}}\mathrm{sin}\:{x} \\ $$$$={x}−\frac{\mathrm{1}}{\mathrm{3}−{D}}\mathrm{sin}\:{x} \\ $$$$={x}−\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{3}+{D}\right)\mathrm{sin}\:{x} \\ $$$$={x}−\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{3sin}\:{x}+\mathrm{cos}\:{x}\right) \\ $$$$\lambda^{\mathrm{3}} −\mathrm{2}\lambda^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\lambda_{\mathrm{1}} =\mathrm{1}\:\:\:\:\:\:\lambda_{\mathrm{2}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\:\:\lambda_{\mathrm{3}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow{y}={C}_{\mathrm{1}} {e}^{{x}} +{C}_{\mathrm{2}} {e}^{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}{x}} +{C}_{\mathrm{3}} {e}^{\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}{x}} +{x}−\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{3sin}\:{x}+\mathrm{cos}\:{x}\right) \\ $$
Commented by mathmax by abdo last updated on 18/Mar/21
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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