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Question Number 136734 by mohammad17 last updated on 25/Mar/21

Answered by Ñï= last updated on 25/Mar/21

Q5::: y_p =(1/(D^2 +3D+12))(e^x cos x−cos x) =e^x (1/(D^2 +5D+13))cos x−(1/(D^2 +3D+12))cos x =e^x (1/(5D+12))cos x−(1/(3D+11))cos x =e^x ((5D−12)/(25D^2 −144))cos x−((3D−11)/(9D^2 −121))cos x =(e^x /(169))(12cos x+5sin x)−(1/(130))(3sin x+11cos x) y=e^(−(3/2)x) (C_1 sin ((√(39))/2)x+C_2 cos ((√(39))/2)x)+(e^x /(169))(12cos x+5sin x)−(1/(130))(3sin x+11cos x) −−−−−−−−−−−−−−−−−−− Q6::: y_p =(1/(D^4 +2D^2 −228D))(2sin x−x^2 −2) =(2/(51985))(228cos x−sin x)+(1/3)∙((x^3 /(228))+((6+6x^2 )/(228^2 ))+((24x)/(228^3 ))+((48)/(228^4 )))−(1/(342)) y=C_1 +C_2 e^(−6x) +(C_3 +C_4 x)e^(6x) +(2/(51985))(228cos x−sin x)+(x^3 /(684))+(x^2 /(25992))+(x/(1481544)) ???

$${Q}\mathrm{5}::: \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{3}{D}+\mathrm{12}}\left({e}^{{x}} \mathrm{cos}\:{x}−\mathrm{cos}\:{x}\right) \\ $$$$={e}^{{x}} \frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{5}{D}+\mathrm{13}}\mathrm{cos}\:{x}−\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{3}{D}+\mathrm{12}}\mathrm{cos}\:{x} \\ $$$$={e}^{{x}} \frac{\mathrm{1}}{\mathrm{5}{D}+\mathrm{12}}\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{3}{D}+\mathrm{11}}\mathrm{cos}\:{x} \\ $$$$={e}^{{x}} \frac{\mathrm{5}{D}−\mathrm{12}}{\mathrm{25}{D}^{\mathrm{2}} −\mathrm{144}}\mathrm{cos}\:{x}−\frac{\mathrm{3}{D}−\mathrm{11}}{\mathrm{9}{D}^{\mathrm{2}} −\mathrm{121}}\mathrm{cos}\:{x} \\ $$$$=\frac{{e}^{{x}} }{\mathrm{169}}\left(\mathrm{12cos}\:{x}+\mathrm{5sin}\:{x}\right)−\frac{\mathrm{1}}{\mathrm{130}}\left(\mathrm{3sin}\:{x}+\mathrm{11cos}\:{x}\right) \\ $$$${y}={e}^{−\frac{\mathrm{3}}{\mathrm{2}}{x}} \left({C}_{\mathrm{1}} \mathrm{sin}\:\frac{\sqrt{\mathrm{39}}}{\mathrm{2}}{x}+{C}_{\mathrm{2}} \mathrm{cos}\:\frac{\sqrt{\mathrm{39}}}{\mathrm{2}}{x}\right)+\frac{{e}^{{x}} }{\mathrm{169}}\left(\mathrm{12cos}\:{x}+\mathrm{5sin}\:{x}\right)−\frac{\mathrm{1}}{\mathrm{130}}\left(\mathrm{3sin}\:{x}+\mathrm{11cos}\:{x}\right) \\ $$$$−−−−−−−−−−−−−−−−−−− \\ $$$${Q}\mathrm{6}::: \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{4}} +\mathrm{2}{D}^{\mathrm{2}} −\mathrm{228}{D}}\left(\mathrm{2sin}\:{x}−{x}^{\mathrm{2}} −\mathrm{2}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{51985}}\left(\mathrm{228cos}\:{x}−\mathrm{sin}\:{x}\right)+\frac{\mathrm{1}}{\mathrm{3}}\centerdot\left(\frac{{x}^{\mathrm{3}} }{\mathrm{228}}+\frac{\mathrm{6}+\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{228}^{\mathrm{2}} }+\frac{\mathrm{24}{x}}{\mathrm{228}^{\mathrm{3}} }+\frac{\mathrm{48}}{\mathrm{228}^{\mathrm{4}} }\right)−\frac{\mathrm{1}}{\mathrm{342}} \\ $$$${y}={C}_{\mathrm{1}} +{C}_{\mathrm{2}} {e}^{−\mathrm{6}{x}} +\left({C}_{\mathrm{3}} +{C}_{\mathrm{4}} {x}\right){e}^{\mathrm{6}{x}} +\frac{\mathrm{2}}{\mathrm{51985}}\left(\mathrm{228cos}\:{x}−\mathrm{sin}\:{x}\right)+\frac{{x}^{\mathrm{3}} }{\mathrm{684}}+\frac{{x}^{\mathrm{2}} }{\mathrm{25992}}+\frac{{x}}{\mathrm{1481544}} \\ $$$$ \\ $$$$??? \\ $$

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