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Question Number 94455 by mhmd last updated on 18/May/20

$$Solve(D^2-2D+1)y={xe}^{x}sinx$$$$pleashelpmesir$$

Commented by mhmd last updated on 19/May/20

$$??$$

Commented by john santu last updated on 19/May/20

$$\mathrm{homogenous}\mathrm{solution}$$$${\lambda }^2-2\lambda +1=0\Rightarrow \lambda =1,1$$$${\mathrm{y}}_{\mathrm{h}}={\mathrm{Ae}}^{\mathrm{x}}+{\mathrm{Bxe}}^{\mathrm{x}}$$

Commented by mhmd last updated on 19/May/20

$$sorysiriwantspecialsolutiontotheequation$$

Answered by Mr.D.N. last updated on 19/May/20

$$({\mathrm{D}}^2-2\mathrm{D}+1)={\mathrm{xe}}^{\mathrm{x}}\sin \mathrm{x}$$$$\mathrm{Auxilairy}\mathrm{Eaquation}\mathrm{we}\mathrm{get},$$$${\mathrm{m}}^2-2\mathrm{m}+1=0,{(\mathrm{m}-1)}^2=0$$$$\mathrm{m}=1,1$$$$\mathrm{Complemenatary}\mathrm{Function}\left(\mathrm{CF}\right)=({\mathrm{C}}_1+{\mathrm{C}}_2\mathrm{x}){\mathrm{e}}^{\mathrm{x}}$$$$\mathrm{Particular}\mathrm{Integral}\left(\mathrm{PI}\right)=\frac{{\mathrm{xe}}^{\mathrm{x}}\sin \mathrm{x}}{{\mathrm{D}}^2-2\mathrm{D}+1}$$$$={\mathrm{e}}^{\mathrm{x}}\frac{1}{{(\mathrm{D}-1)}^2}\mathrm{x}\mathrm{sinx}$$$$={\mathrm{e}}^{\mathrm{x}}\frac{1}{{(\mathrm{D}+1-1)}^2}\mathrm{x}\sin \mathrm{x}$$$$={\mathrm{e}}^{\mathrm{x}}\frac{\mathrm{x}\sin \mathrm{x}}{{\mathrm{D}}^2}={\mathrm{e}}^{\mathrm{x}}\int (\int \mathrm{x}\sin \mathrm{x}\mathrm{dx})\mathrm{dx}$$$$\mathrm{x}(-\cos \mathrm{x})-\int 1.(-\cos \mathrm{x})\mathrm{dx}$$$$-\mathrm{xcos}\mathrm{x}+\sin \mathrm{x}$$$$-\int \mathrm{x}\cos \mathrm{x}+\int \sin \mathrm{x}$$$$-[\mathrm{x}\cos \mathrm{x}-\int \sin \mathrm{xdx}]-\cos \mathrm{x}$$$$-\mathrm{x}\cos \mathrm{x}-\cos \mathrm{x}-\cos \mathrm{x}$$$$-\mathrm{xcos}\mathrm{x}-2\cos \mathrm{x}$$$$\mathrm{PI}=-{\mathrm{e}}^{\mathrm{x}}(\mathrm{x}\cos \mathrm{x}+2\cos \mathrm{x})$$$$\mathrm{y}=\mathrm{CF}+\mathrm{PI}$$$$\mathrm{y}=({\mathrm{C}}_1+{\mathrm{C}}_2\mathrm{x}){\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}(\mathrm{x}\cos \mathrm{x}+2\cos \mathrm{x}).$$

Commented by mhmd last updated on 19/May/20

$$thankyousir$$

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