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Question Number 135402 by mohammad17 last updated on 12/Mar/21

Answered by Ñï= last updated on 13/Mar/21

y_p =(1/(D^2 −2D+2))(4xe^x cos x+xe^(−x) +x^2 +1) 1>>(1/(D^2 −2D+2))4xe^x cos x=2(1/(D^2 −2D+2))e^x (e^(ix) +e^(−ix) )x =2{e^((1+i)x) (1/((D+1+i)^2 −2(D+1+i)+2))x+e^((1−i)x) (1/((D+1−i)^2 −2(D+1−i)+2))x} =2{e^((1+i)x) (1/((D+2i)D))x+e^((1−i)x) (1/((D−2i)D))x}={e^((1+i)x) (1/(D+2i))x^2 +e^((1−i)x) (1/(D−2i))x^2 } =e^((1+i)x) (1/(2i(1+(D/(2i)))))x^2 −e^((1−i)x) (1/(2i(1−(D/(2i)))))x^2 =(1/(2i))e^((1+i)x) {1+(−(D/(2i)))+(−(D/(2i)))^2 }x^2 −(e^((1−i)x) /(2i)){1+((D/(2i)))+((D/(2i)))^2 }x^2 =(1/(2i))e^((1+i)x) {x^2 −(x/i)−(1/2)}−(1/(2i))e^((1−i)x) {x^2 +(x/i)−(1/2)} =e^((1+i)x) {(x^2 /(2i))+(x/2)−(1/(4i))}+e^((1−i)x) {−(x^2 /(2i))+(x/2)+(1/(4i))} =e^x {e^(ix) (x^2 /(2i))+e^(ix) (x/2)−e^(ix) (1/(4i))−e^(−ix) (x^2 /(2i))+e^(−ix) (x/2)+e^(−ix) (1/(4i))} =e^x (xcos x−(1/2)sin x+x^2 sin x) 2>>(1/(D^2 −2D+2))xe^(−x) =e^(−x) (1/((D−1)^2 −2(D−1)+2))x =e^(−x) (1/(D^2 −4D+5))x=e^(−x) (1/(5(1+(D^2 /5)−((4D)/5))))x =(e^(−x) /5)(1+((4D)/5)−(D^2 /5))x=(e^(−x) /5)(x+(4/5)) 3>>(1/(D^2 −2D+2))(x^2 +1)=(1/(2(1−D+(D^2 /2))))(x^2 +1) =(1/2)(1+D−(D^2 /2)+(D−(D^2 /2))^2 +...)(x^2 +1) =(1/2)(1+D+(D^2 /2)+...)(x^2 +1)=(1/2)(x^2 +2x+2) y_p =e^x (xcos x−(1/2)sin x+x^2 sin x)+(1/5)(x+(4/5))e^(−x) +(1/2)(x^2 +2x) λ^2 −2λ+2=0 ⇒λ=1±i ⇒y_h =e^x (C_1 ′sin x+C_2 cos x) y=y_h +y_p =e^x (C_1 sin x+C_2 cos x+xcos x+x^2 sin x)+(1/5)(x+(4/5))e^(−x) +(1/2)(x^2 +2x+2)

$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{2}}\left(\mathrm{4}{xe}^{{x}} \mathrm{co}{s}\:{x}+{xe}^{−{x}} +{x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{1}>>\frac{\mathrm{1}}{{D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{2}}\mathrm{4}{xe}^{{x}} {cos}\:{x}=\mathrm{2}\frac{\mathrm{1}}{{D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{2}}{e}^{{x}} \left({e}^{{ix}} +{e}^{−{ix}} \right){x} \\ $$$$=\mathrm{2}\left\{{e}^{\left(\mathrm{1}+{i}\right){x}} \frac{\mathrm{1}}{\left({D}+\mathrm{1}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({D}+\mathrm{1}+{i}\right)+\mathrm{2}}{x}+{e}^{\left(\mathrm{1}−{i}\right){x}} \frac{\mathrm{1}}{\left({D}+\mathrm{1}−{i}\right)^{\mathrm{2}} −\mathrm{2}\left({D}+\mathrm{1}−{i}\right)+\mathrm{2}}{x}\right\} \\ $$$$=\mathrm{2}\left\{{e}^{\left(\mathrm{1}+{i}\right){x}} \frac{\mathrm{1}}{\left({D}+\mathrm{2}{i}\right){D}}{x}+{e}^{\left(\mathrm{1}−{i}\right){x}} \frac{\mathrm{1}}{\left({D}−\mathrm{2}{i}\right){D}}{x}\right\}=\left\{{e}^{\left(\mathrm{1}+{i}\right){x}} \frac{\mathrm{1}}{{D}+\mathrm{2}{i}}{x}^{\mathrm{2}} +{e}^{\left(\mathrm{1}−{i}\right){x}} \frac{\mathrm{1}}{{D}−\mathrm{2}{i}}{x}^{\mathrm{2}} \right\} \\ $$$$={e}^{\left(\mathrm{1}+{i}\right){x}} \frac{\mathrm{1}}{\mathrm{2}{i}\left(\mathrm{1}+\frac{{D}}{\mathrm{2}{i}}\right)}{x}^{\mathrm{2}} −{e}^{\left(\mathrm{1}−{i}\right){x}} \frac{\mathrm{1}}{\mathrm{2}{i}\left(\mathrm{1}−\frac{{D}}{\mathrm{2}{i}}\right)}{x}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{e}^{\left(\mathrm{1}+{i}\right){x}} \left\{\mathrm{1}+\left(−\frac{{D}}{\mathrm{2}{i}}\right)+\left(−\frac{{D}}{\mathrm{2}{i}}\right)^{\mathrm{2}} \right\}{x}^{\mathrm{2}} −\frac{{e}^{\left(\mathrm{1}−{i}\right){x}} }{\mathrm{2}{i}}\left\{\mathrm{1}+\left(\frac{{D}}{\mathrm{2}{i}}\right)+\left(\frac{{D}}{\mathrm{2}{i}}\right)^{\mathrm{2}} \right\}{x}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{e}^{\left(\mathrm{1}+{i}\right){x}} \left\{{x}^{\mathrm{2}} −\frac{{x}}{{i}}−\frac{\mathrm{1}}{\mathrm{2}}\right\}−\frac{\mathrm{1}}{\mathrm{2}{i}}{e}^{\left(\mathrm{1}−{i}\right){x}} \left\{{x}^{\mathrm{2}} +\frac{{x}}{{i}}−\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$={e}^{\left(\mathrm{1}+{i}\right){x}} \left\{\frac{{x}^{\mathrm{2}} }{\mathrm{2}{i}}+\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}{i}}\right\}+{e}^{\left(\mathrm{1}−{i}\right){x}} \left\{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}{i}}+\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}{i}}\right\} \\ $$$$={e}^{{x}} \left\{{e}^{{ix}} \frac{{x}^{\mathrm{2}} }{\mathrm{2}{i}}+{e}^{{ix}} \frac{{x}}{\mathrm{2}}−{e}^{{ix}} \frac{\mathrm{1}}{\mathrm{4}{i}}−{e}^{−{ix}} \frac{{x}^{\mathrm{2}} }{\mathrm{2}{i}}+{e}^{−{ix}} \frac{{x}}{\mathrm{2}}+{e}^{−{ix}} \frac{\mathrm{1}}{\mathrm{4}{i}}\right\} \\ $$$$={e}^{{x}} \left({x}\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\:{x}+{x}^{\mathrm{2}} {sin}\:{x}\right) \\ $$$$\mathrm{2}>>\frac{\mathrm{1}}{{D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{2}}{xe}^{−{x}} ={e}^{−{x}} \frac{\mathrm{1}}{\left({D}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({D}−\mathrm{1}\right)+\mathrm{2}}{x} \\ $$$$={e}^{−{x}} \frac{\mathrm{1}}{{D}^{\mathrm{2}} −\mathrm{4}{D}+\mathrm{5}}{x}={e}^{−{x}} \frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{1}+\frac{{D}^{\mathrm{2}} }{\mathrm{5}}−\frac{\mathrm{4}{D}}{\mathrm{5}}\right)}{x} \\ $$$$=\frac{{e}^{−{x}} }{\mathrm{5}}\left(\mathrm{1}+\frac{\mathrm{4}{D}}{\mathrm{5}}−\frac{{D}^{\mathrm{2}} }{\mathrm{5}}\right){x}=\frac{{e}^{−{x}} }{\mathrm{5}}\left({x}+\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\mathrm{3}>>\frac{\mathrm{1}}{{D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−{D}+\frac{{D}^{\mathrm{2}} }{\mathrm{2}}\right)}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{D}−\frac{{D}^{\mathrm{2}} }{\mathrm{2}}+\left({D}−\frac{{D}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} +...\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{D}+\frac{{D}^{\mathrm{2}} }{\mathrm{2}}+...\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right) \\ $$$${y}_{{p}} ={e}^{{x}} \left({x}\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{x}+{x}^{\mathrm{2}} {sin}\:{x}\right)+\frac{\mathrm{1}}{\mathrm{5}}\left({x}+\frac{\mathrm{4}}{\mathrm{5}}\right){e}^{−{x}} +\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right) \\ $$$$\lambda^{\mathrm{2}} −\mathrm{2}\lambda+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\lambda=\mathrm{1}\pm{i} \\ $$$$\Rightarrow{y}_{{h}} ={e}^{{x}} \left({C}_{\mathrm{1}} '\mathrm{sin}\:{x}+{C}_{\mathrm{2}} \mathrm{cos}\:{x}\right) \\ $$$${y}={y}_{{h}} +{y}_{{p}} ={e}^{{x}} \left({C}_{\mathrm{1}} \mathrm{sin}\:{x}+{C}_{\mathrm{2}} \mathrm{cos}\:{x}+{x}\mathrm{cos}\:{x}+{x}^{\mathrm{2}} \mathrm{sin}\:{x}\right)+\frac{\mathrm{1}}{\mathrm{5}}\left({x}+\frac{\mathrm{4}}{\mathrm{5}}\right){e}^{−{x}} +\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right) \\ $$

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