Question Number 131488 by Ahmed1hamouda last updated on 05/Feb/21
Answered by Ñï= last updated on 18/Feb/21
![y′′+2y′+5y=6e^(2x) +xsin^2 x+e^x cos2x y_p =(1/(D^2 +2D+5))(6e^(2x) +xsin^2 x+e^x cos2x) =(6/(2^2 +2×2+5))e^(2x) +e^x (D^2 −2D+5)(1/((D^2 +5)^2 −4D^2 ))cos2x+(1/(D^2 +2D+5))xsin^2 x =(6/(13))e^(2x) +e^x (1−2D)cos2x+(1/(D^2 +2D+5))Im(e^(2ix) x) =(6/(13))e^(2x) +e^x (cos2x+4sin2x)+Im((1/(D^2 +2D+5))e^(2ix) x) =(6/(13))e^(2x) +e^x (cos2x+4sin2x)+Im(e^(2ix) (1/((D+2i)^2 +4i+5))x) =(6/(13))e^(2x) +e^x (cos2x+4sin2x)+Im(e^(2ix) (1/(4i+1))∙(1/((D^2 /(4i+1))+((4iD)/(4i+1))+1))x) =(6/(13))e^(2x) +e^x (cos2x+4sin2x)+Im[e^(2ix) (1/(4i+1))∙(x−((4i)/(4i+1)))] =(6/(13))e^(2x) +e^x (cos2x+4sin2x)−((4x)/(15))cos2x−(x/(15))sin2x−((16)/(15))sin2x−(4/(15))cos2x λ^2 +2λ+5=0⇒λ_1 =−1+2i,λ=−1−2i y=C_1 e^(−x) sin2x+C_2 e^(−x) cos2x+(6/(13))e^(2x) +e^x (cos2x+4sin2x)−((4x)/(15))cos2x−(x/(15))sin2x−((16)/(15))sin2x−(4/(15))cos2x](https://www.tinkutara.com/question/Q133042.png)
$${y}''+\mathrm{2}{y}'+\mathrm{5}{y}=\mathrm{6}{e}^{\mathrm{2}{x}} +{xsin}^{\mathrm{2}} {x}+{e}^{{x}} {cos}\mathrm{2}{x} \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{2}{D}+\mathrm{5}}\left(\mathrm{6}{e}^{\mathrm{2}{x}} +{xsin}^{\mathrm{2}} {x}+{e}^{{x}} {cos}\mathrm{2}{x}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{2}} +\mathrm{2}×\mathrm{2}+\mathrm{5}}{e}^{\mathrm{2}{x}} +{e}^{{x}} \left({D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{5}\right)\frac{\mathrm{1}}{\left({D}^{\mathrm{2}} +\mathrm{5}\right)^{\mathrm{2}} −\mathrm{4}{D}^{\mathrm{2}} }{cos}\mathrm{2}{x}+\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{2}{D}+\mathrm{5}}{xsin}^{\mathrm{2}} {x} \\ $$$$\:\:\:\:=\frac{\mathrm{6}}{\mathrm{13}}{e}^{\mathrm{2}{x}} +{e}^{{x}} \left(\mathrm{1}−\mathrm{2}{D}\right){cos}\mathrm{2}{x}+\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{2}{D}+\mathrm{5}}{Im}\left({e}^{\mathrm{2}{ix}} {x}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{6}}{\mathrm{13}}{e}^{\mathrm{2}{x}} +{e}^{{x}} \left({cos}\mathrm{2}{x}+\mathrm{4}{sin}\mathrm{2}{x}\right)+{Im}\left(\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{2}{D}+\mathrm{5}}{e}^{\mathrm{2}{ix}} {x}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{6}}{\mathrm{13}}{e}^{\mathrm{2}{x}} +{e}^{{x}} \left({cos}\mathrm{2}{x}+\mathrm{4}{sin}\mathrm{2}{x}\right)+{Im}\left({e}^{\mathrm{2}{ix}} \frac{\mathrm{1}}{\left({D}+\mathrm{2}{i}\right)^{\mathrm{2}} +\mathrm{4}{i}+\mathrm{5}}{x}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{6}}{\mathrm{13}}{e}^{\mathrm{2}{x}} +{e}^{{x}} \left({cos}\mathrm{2}{x}+\mathrm{4}{sin}\mathrm{2}{x}\right)+{Im}\left({e}^{\mathrm{2}{ix}} \frac{\mathrm{1}}{\mathrm{4}{i}+\mathrm{1}}\centerdot\frac{\mathrm{1}}{\frac{{D}^{\mathrm{2}} }{\mathrm{4}{i}+\mathrm{1}}+\frac{\mathrm{4}{iD}}{\mathrm{4}{i}+\mathrm{1}}+\mathrm{1}}{x}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{6}}{\mathrm{13}}{e}^{\mathrm{2}{x}} +{e}^{{x}} \left({cos}\mathrm{2}{x}+\mathrm{4}{sin}\mathrm{2}{x}\right)+{Im}\left[{e}^{\mathrm{2}{ix}} \frac{\mathrm{1}}{\mathrm{4}{i}+\mathrm{1}}\centerdot\left({x}−\frac{\mathrm{4}{i}}{\mathrm{4}{i}+\mathrm{1}}\right)\right] \\ $$$$\:\:\:\:=\frac{\mathrm{6}}{\mathrm{13}}{e}^{\mathrm{2}{x}} +{e}^{{x}} \left({cos}\mathrm{2}{x}+\mathrm{4}{sin}\mathrm{2}{x}\right)−\frac{\mathrm{4}{x}}{\mathrm{15}}{cos}\mathrm{2}{x}−\frac{{x}}{\mathrm{15}}{sin}\mathrm{2}{x}−\frac{\mathrm{16}}{\mathrm{15}}{sin}\mathrm{2}{x}−\frac{\mathrm{4}}{\mathrm{15}}{cos}\mathrm{2}{x} \\ $$$$\lambda^{\mathrm{2}} +\mathrm{2}\lambda+\mathrm{5}=\mathrm{0}\Rightarrow\lambda_{\mathrm{1}} =−\mathrm{1}+\mathrm{2}{i},\lambda=−\mathrm{1}−\mathrm{2}{i} \\ $$$${y}={C}_{\mathrm{1}} {e}^{−{x}} {sin}\mathrm{2}{x}+{C}_{\mathrm{2}} {e}^{−{x}} {cos}\mathrm{2}{x}+\frac{\mathrm{6}}{\mathrm{13}}{e}^{\mathrm{2}{x}} +{e}^{{x}} \left({cos}\mathrm{2}{x}+\mathrm{4}{sin}\mathrm{2}{x}\right)−\frac{\mathrm{4}{x}}{\mathrm{15}}{cos}\mathrm{2}{x}−\frac{{x}}{\mathrm{15}}{sin}\mathrm{2}{x}−\frac{\mathrm{16}}{\mathrm{15}}{sin}\mathrm{2}{x}−\frac{\mathrm{4}}{\mathrm{15}}{cos}\mathrm{2}{x} \\ $$