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Question Number 98106 by bobhans last updated on 11/Jun/20

$${\mathrm{y}}^{″}+\mathrm{y}=\cos 3\mathrm{x}-2\sin 3\mathrm{x}$$

Answered by bemath last updated on 11/Jun/20

$$\mathrm{homogenous}\mathrm{solution}$$$${\lambda }^2+1=0\Rightarrow \lambda =\pm i$$$${\mathrm{y}}_{\mathrm{h}}=\mathrm{Acos}\mathrm{x}+\mathrm{Bsin}\mathrm{x}$$$$\mathrm{particular}\mathrm{solution}$$$${\mathrm{y}}_{\mathrm{p}}=a\cos 3x+b\sin 3x$$$${\mathrm{y}}_{\mathrm{p}}^{\prime }=-3a\sin 3x+3b\cos 3x$$$${\mathrm{y}}_{\mathrm{p}}^{″}=-9\mathrm{acos}3x-9b\sin 3x$$$$\iff (-9\mathrm{acos}3\mathrm{x}-9\mathrm{bsin}3\mathrm{x})+(\mathrm{acos}3\mathrm{x}+\mathrm{bsin}3\mathrm{x})=$$$$\cos 3\mathrm{x}-2\sin 3\mathrm{x}$$$$\iff -8\mathrm{acos}3\mathrm{x}-8\mathrm{bsin}3\mathrm{x}=$$$$\cos 3\mathrm{x}-2\sin 3\mathrm{x}$$$$\mathrm{we}\mathrm{get}\{\begin{array}{l}-8\mathrm{a}=1\to \mathrm{a}=-\frac{1}{8}\\ -8\mathrm{b}=-2\to \mathrm{b}=\frac{1}{4}\end{array}$$$$\mathrm{then}{\mathrm{y}}_{\mathrm{p}}=-\frac{1}{8}\cos 3\mathrm{x}+\frac{1}{4}\sin 3\mathrm{x}$$$$\mathrm{generall}\mathrm{solution}$$$${\mathrm{y}}_{\mathrm{c}}=\mathrm{Acos}\mathrm{x}+\mathrm{Bsin}\mathrm{x}-\frac{1}{8}\cos 3\mathrm{x}+\frac{1}{4}\sin 3\mathrm{x}$$

Answered by mathmax by abdo last updated on 11/Jun/20

$$\left(\mathrm{eh}\right)\to {\mathrm{y}}^{{}^{″}}+\mathrm{y}=0\to {\mathrm{r}}^2+1=0\Rightarrow \mathrm{r}=\stackrel{-}{+}\mathrm{i}\Rightarrow {\mathrm{y}}_{\mathrm{h}}=\mathrm{acosx}+\mathrm{bsinx}$$$${\mathrm{y}}_{\mathrm{p}}\mathrm{is}\mathrm{at}\mathrm{form}{\mathrm{y}}_{\mathrm{p}}=\alpha \cos \left(3\mathrm{x}\right)+\beta \sin \left(3\mathrm{x}\right)$$$${\mathrm{y}}^{{}^{\prime }}=-3\alpha \sin \left(3\mathrm{x}\right)+3\beta \cos \left(3\mathrm{x}\right)\mathrm{and}{\mathrm{y}}^{{}^{″}}=-9\alpha \cos \left(3\mathrm{x}\right)-9\beta \sin \left(3\mathrm{x}\right)$$$$\left(\mathrm{e}\right)\Rightarrow -9\alpha \cos \left(3\mathrm{x}\right)+9\beta \sin \left(3\mathrm{x}\right)+\alpha \cos \left(3\mathrm{x}\right)+\beta \sin \left(3\mathrm{x}\right)=\cos \left(3\mathrm{x}\right)-2\sin \left(3\mathrm{x}\right)\Rightarrow$$$$-8\alpha \cos \left(3\mathrm{x}\right)+10\beta \sin \left(3\mathrm{x}\right)=\cos \left(3\mathrm{x}\right)-2\sin \left(3\mathrm{x}\right)\Rightarrow \{\begin{array}{l}-8\alpha =1\\ 10\beta =-2\end{array}$$$$\{\begin{array}{l}\alpha =-\frac{1}{8}\mathrm{the}\mathrm{genearal}\mathrm{solution}\mathrm{is}\mathrm{y}=\mathrm{acosx}+\mathrm{bsinx}-\frac{1}{8}\cos \left(3\mathrm{x}\right)-\frac{1}{5}\sin \left(3\mathrm{x}\right)\\ \beta =-\frac{1}{5}\end{array}$$

Commented by john santu last updated on 12/Jun/20

$$-9\beta +\beta =-2$$$$-8\beta =-2\Rightarrow \beta =\frac{1}{4}$$

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