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Question Number 146260 by mathmax by abdo last updated on 12/Jul/21

solve y^(′′) −2y^′ +y =e^(−x) sinx

$$\mathrm{solve}\:\mathrm{y}^{''} −\mathrm{2y}^{'} \:+\mathrm{y}\:=\mathrm{e}^{−\mathrm{x}} \mathrm{sinx} \\ $$

Answered by qaz last updated on 12/Jul/21

y_p =(1/(D^2 −2D+1))e^(−x) sin x =e^x (1/D^2 )e^(−2x) sin x =e^x ∫∫e^(−2x) sin xdxdx ...... =(1/(25))e^(−x) (4cos x+3sin x) −−−−−−−−−− y_p =e^(−x) (1/((D−2)^2 ))sin x =e^(−x) (((D+2)^2 )/((D^2 −4)^2 ))sin x =e^(−x) ((−1^2 +4D+4)/((−1^2 −4)^2 ))sin x =(1/(25))e^(−x) (4cos x+3sin x) ⇒ y=e^x (C_1 +C_2 x)+(1/(25))e^(−x) (4cos x+3sin x)

$$\mathrm{y}_{\mathrm{p}} =\frac{\mathrm{1}}{\mathrm{D}^{\mathrm{2}} −\mathrm{2D}+\mathrm{1}}\mathrm{e}^{−\mathrm{x}} \mathrm{sin}\:\mathrm{x} \\ $$$$=\mathrm{e}^{\mathrm{x}} \frac{\mathrm{1}}{\mathrm{D}^{\mathrm{2}} }\mathrm{e}^{−\mathrm{2x}} \mathrm{sin}\:\mathrm{x} \\ $$$$=\mathrm{e}^{\mathrm{x}} \int\int\mathrm{e}^{−\mathrm{2x}} \mathrm{sin}\:\mathrm{xdxdx} \\ $$$$...... \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\mathrm{e}^{−\mathrm{x}} \left(\mathrm{4cos}\:\mathrm{x}+\mathrm{3sin}\:\mathrm{x}\right) \\ $$$$−−−−−−−−−− \\ $$$$\mathrm{y}_{\mathrm{p}} =\mathrm{e}^{−\mathrm{x}} \frac{\mathrm{1}}{\left(\mathrm{D}−\mathrm{2}\right)^{\mathrm{2}} }\mathrm{sin}\:\mathrm{x} \\ $$$$=\mathrm{e}^{−\mathrm{x}} \frac{\left(\mathrm{D}+\mathrm{2}\right)^{\mathrm{2}} }{\left(\mathrm{D}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} }\mathrm{sin}\:\mathrm{x} \\ $$$$=\mathrm{e}^{−\mathrm{x}} \frac{−\mathrm{1}^{\mathrm{2}} +\mathrm{4D}+\mathrm{4}}{\left(−\mathrm{1}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} }\mathrm{sin}\:\mathrm{x} \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\mathrm{e}^{−\mathrm{x}} \left(\mathrm{4cos}\:\mathrm{x}+\mathrm{3sin}\:\mathrm{x}\right) \\ $$$$\Rightarrow\:\mathrm{y}=\mathrm{e}^{\mathrm{x}} \left(\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} \mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{25}}\mathrm{e}^{−\mathrm{x}} \left(\mathrm{4cos}\:\mathrm{x}+\mathrm{3sin}\:\mathrm{x}\right) \\ $$

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