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Question Number 100344 by bemath last updated on 26/Jun/20

$$\mathrm{solve}{\mathrm{y}}^{″}+\mathrm{y}=\sin \mathrm{x}$$

Answered by bobhans last updated on 26/Jun/20

$$\mathrm{homogenous}\mathrm{solution}$$$${\eta }^2+1=0\Rightarrow \eta =\pm \mathrm{i}$$$${\mathrm{y}}_{\mathrm{h}}=\mathrm{Asin}\mathrm{x}+\mathrm{Bcos}\mathrm{x}$$$$\mathrm{particular}\mathrm{solution}$$$${\mathrm{y}}_{\mathrm{p}}=x(A\sin \mathrm{x}+\mathrm{Bcos}\mathrm{x})$$$${\mathrm{y}}_{\mathrm{p}}^{\prime }=\mathrm{Asin}\mathrm{x}+\mathrm{Bcos}\mathrm{x}+x(A\cos \mathrm{x}-\mathrm{Bsin}\mathrm{x})$$$${\mathrm{y}}_{\mathrm{p}}^{″}=2\mathrm{Acos}\mathrm{x}-2\mathrm{Bsin}\mathrm{x}+x(-A\sin \mathrm{x}-\mathrm{Bcos}\mathrm{x})$$$$\mathrm{A}=0\wedge \mathrm{B}=\frac{1}{2}$$$${\mathrm{y}}_{\mathrm{p}}=\frac{1}{2}x\cos x$$$$\mathbb{G}\mathrm{enerall}\mathrm{solution}$$$${\mathrm{y}}_g=A\sin \mathrm{x}+\mathrm{Bcos}\mathrm{x}+\frac{1}{2}\mathrm{x}\cos \mathrm{x}$$

Answered by mathmax by abdo last updated on 26/Jun/20

$${\mathrm{y}}^{{}^{″}}+\mathrm{y}=\mathrm{sinx}\left(\mathrm{he}\right)\to {\mathrm{r}}^2+1=0\Rightarrow \mathrm{r}=\stackrel{-}{+}\mathrm{i}\Rightarrow {\mathrm{y}}_{\mathrm{h}}={\mathrm{ae}}^{\mathrm{ix}}+{\mathrm{be}}^{-\mathrm{ix}}=\alpha \mathrm{cosx}+\beta \mathrm{sinx}$$$$=\alpha {\mathrm{u}}_1+\beta {\mathrm{u}}_2$$$$\mathrm{W}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}\mathrm{cosx}\mathrm{sinx}\\ -\mathrm{sinx}\mathrm{cosx}\end{array}\right|=1$$$${\mathrm{W}}_1=\left|\begin{array}{c}0\mathrm{sinx}\\ \mathrm{sinx}\mathrm{cosx}\end{array}\right|=-{\sin }^2\mathrm{x}$$$${\mathrm{W}}_2=\left|\begin{array}{c}\mathrm{cosx}0\\ -\mathrm{sinx}\mathrm{sinx}\end{array}\right|=\mathrm{cosx}\mathrm{sinx}$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=-\int \frac{{\sin }^2\mathrm{x}}{1}\mathrm{dx}=-\int \frac{1-\cos \left(2\mathrm{x}\right)}{2}=-\frac{\mathrm{x}}{2}+\frac{1}{4}\sin \left(2\mathrm{x}\right)$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=\int \mathrm{cosx}\mathrm{sinx}\mathrm{dx}=\frac{1}{2}\int \sin \left(2\mathrm{x}\right)\mathrm{dx}=-\frac{1}{4}\cos \left(2\mathrm{x}\right)\Rightarrow$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2=\mathrm{cosx}(-\frac{\mathrm{x}}{2}+\frac{1}{4}\sin \left(2\mathrm{x}\right))+\mathrm{sinx}(-\frac{1}{4}\cos \left(2\mathrm{x}\right))$$$$=-\frac{\mathrm{x}}{2}\mathrm{cosx}+\frac{1}{4}\mathrm{cosx}\sin \left(2\mathrm{x}\right)-\frac{1}{4}\mathrm{sinx}\cos \left(2\mathrm{x}\right)$$$$=-\frac{\mathrm{x}}{2}\mathrm{cosx}+\frac{1}{2}{\cos }^2\mathrm{x}\mathrm{sinx}-\frac{1}{4}\mathrm{sinx}({2\cos }^2\mathrm{x}-1)$$$$=-\frac{\mathrm{x}}{2}\mathrm{cosx}+\frac{1}{4}\mathrm{sinx}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}$$$$\mathrm{y}={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}=\alpha \mathrm{cosx}+\beta \mathrm{sinx}-\frac{\mathrm{x}}{2}\mathrm{cosx}+\frac{1}{4}\mathrm{sinx}$$$$$$

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