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