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Question Number 126854 by sdfg last updated on 24/Dec/20

Answered by mathmax by abdo last updated on 24/Dec/20

$$\mathrm{h}\to {\mathrm{r}}^2+{\mathrm{k}}^2=0\Rightarrow {\mathrm{r}}^2=-{\mathrm{k}}^2\Rightarrow \mathrm{r}=\stackrel{-}{+}\mathrm{ik}\Rightarrow {\mathrm{y}}_{\mathrm{h}}={\mathrm{ae}}^{\mathrm{ikx}}+{\mathrm{be}}^{-\mathrm{ikx}}$$$$=\alpha \cos \left(\mathrm{kx}\right)+\beta \sin \left(\mathrm{kx}\right)=\alpha {\mathrm{u}}_1+\beta {\mathrm{u}}_2$$$$\mathrm{w}({\mathrm{u}}_1,{\mathrm{u}}_2)=\left|\begin{array}{c}\cos \left(\mathrm{kx}\right)\sin \left(\mathrm{kx}\right)\\ -\mathrm{ksin}\left(\mathrm{kx}\right)\mathrm{kcos}\left(\mathrm{x}\right)\end{array}\right|=\mathrm{k}(\mathrm{we}\mathrm{suppose}\mathrm{k}\ne \mathrm{o})$$$${\mathrm{w}}_1=\left|\begin{array}{c}\mathrm{o}\sin \left(\mathrm{kx}\right)\\ \mathrm{g}\left(\mathrm{x}\right)\mathrm{kcos}\left(\mathrm{kx}\right)\end{array}\right|=-\sin \left(\mathrm{kx}\right)\mathrm{g}\left(\mathrm{x}\right)$$$${\mathrm{w}}_2=\left|\begin{array}{c}\cos \left(\mathrm{kx}\right)0\\ -\mathrm{ksin}\left(\mathrm{kx}\right)\mathrm{g}\left(\mathrm{x}\right)\end{array}\right|=\cos \left(\mathrm{kx}\right)\mathrm{g}\left(\mathrm{x}\right)$$$${\mathrm{v}}_1=\int \frac{{\mathrm{w}}_1}{\mathrm{w}}\mathrm{dx}=-\frac{1}{\mathrm{k}}\int \sin \left(\mathrm{kx}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}$$$${\mathrm{v}}_2=\int \frac{{\mathrm{w}}_2}{\mathrm{w}}\mathrm{dx}=\frac{1}{\mathrm{k}}\int \cos \left(\mathrm{kx}\right)\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx}\Rightarrow$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{u}}_1{\mathrm{v}}_1+{\mathrm{u}}_2{\mathrm{v}}_2=-\frac{1}{\mathrm{k}}\cos \left(\mathrm{kx}\right){\int }_0^{\mathrm{x}}\sin \left(\mathrm{kt}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}+\frac{1}{\mathrm{k}}\sin \left(\mathrm{kx}\right){\int }_0^{\mathrm{x}}\cos \left(\mathrm{kt}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$$$$\Rightarrow \mathrm{y}\left(\mathrm{x}\right)={\mathrm{y}}_{\mathrm{h}}+{\mathrm{y}}_{\mathrm{p}}$$$$=\alpha \cos \left(\mathrm{kx}\right)+\beta \sin \left(\mathrm{kx}\right)-\frac{1}{\mathrm{k}}\cos \left(\mathrm{kx}\right){\int }_0^{\mathrm{x}}\sin \left(\mathrm{kt}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}+\frac{1}{\mathrm{k}}\sin \left(\mathrm{kx}\right){\int }_0^{\mathrm{x}}\cos \left(\mathrm{kt}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$$$$\mathrm{y}\left(0\right)=\mathrm{a}\Rightarrow \alpha =\mathrm{a}$$$${\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)=-\alpha \mathrm{ksin}\left(\mathrm{kx}\right)+\beta \mathrm{kcos}\left(\mathrm{kx}\right)-\frac{1}{\mathrm{k}}\{\mathrm{ksin}\left(\mathrm{kx}\right){\int }_0^{\mathrm{x}}\sin \left(\mathrm{kt}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$$$$+\cos \left(\mathrm{kx}\right)\sin \left(\mathrm{kx}\right)\mathrm{g}\left(\mathrm{x}\right)\}+\frac{1}{\mathrm{k}}\{\mathrm{kcos}\left(\mathrm{kx}\right){\int }_0^{\mathrm{x}}\cos \left(\mathrm{kt}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$$$$+\sin \left(\mathrm{kx}\right)\cos \left(\mathrm{kx}\right)\mathrm{g}\left(\mathrm{x}\right)\}$$$$=-\alpha \mathrm{k}\sin \left(\mathrm{kx}\right)+\beta \mathrm{k}\cos \left(\mathrm{kx}\right)-\sin \left(\mathrm{kx}\right){\int }_0^{\mathrm{x}}\sin \left(\mathrm{kt}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$$$$+\cos \left(\mathrm{kx}\right){\int }_0^{\mathrm{x}}\cos \left(\mathrm{kt}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$$$${\mathrm{y}}^{{}^{\prime }}\left(0\right)=\mathrm{b}\Rightarrow \beta \mathrm{k}=\mathrm{b}\Rightarrow \beta =\frac{\mathrm{b}}{\mathrm{k}}\mathrm{so}\mathrm{y}\left(\mathrm{x}\right)\mathrm{is}\mathrm{known}$$

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