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Question Number 145773 by help last updated on 07/Jul/21

Answered by mathmax by abdo last updated on 08/Jul/21

$$\to {4\mathrm{r}}^2+24\mathrm{r}+37=0$$$${\mathrm{\Delta }}^{{}^{\prime }}={12}^2-4\times 37=144-148=-4\Rightarrow$$$${\mathrm{r}}_1=\frac{-12+2\mathrm{i}}{4}=-3+\frac{1}{2}\mathrm{i}\mathrm{and}{\mathrm{r}}_2=\frac{-12-2\mathrm{i}}{4}=-3-\frac{1}{2}\mathrm{i}$$$$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{ae}}^{{\mathrm{r}}_1\mathrm{x}}+{\mathrm{be}}^{{\mathrm{r}}_2\mathrm{x}}={\mathrm{ae}}^{(-3+\frac{\mathrm{i}}{2})\mathrm{x}}+\mathrm{b}{\mathrm{e}}^{(-3+\frac{\mathrm{i}}{2})\mathrm{x}}$$$$={\mathrm{e}}^{-3\mathrm{x}}(\alpha \cos \left(\frac{\mathrm{x}}{2}\right)+\beta \sin \left(\frac{\mathrm{x}}{2}\right))$$$$\mathrm{y}\left(\pi \right)=1\Rightarrow {\mathrm{e}}^{-3\pi }(\alpha 0+\beta )=1\Rightarrow \beta ={\mathrm{e}}^{3\pi }$$$${\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)=-{3\mathrm{e}}^{-3\mathrm{x}}(\alpha \cos \left(\frac{\mathrm{x}}{2}\right)+\beta \sin \left(\frac{\mathrm{x}}{2}\right))+{\mathrm{e}}^{-3\mathrm{x}}(-\frac{\alpha }{2}\sin \left(\frac{\mathrm{x}}{2}\right)+\frac{\beta }{2}\cos \left(\frac{\mathrm{x}}{2}\right))$$$${\mathrm{y}}^{{}^{\prime }}\left(\pi \right)=0\Rightarrow -{3\mathrm{e}}^{-3\pi }\left(\beta \right)+{\mathrm{e}}^{-3\pi }(-\frac{\alpha }{2})=0\Rightarrow$$$$-3-\frac{{\mathrm{e}}^{-3\pi }}{2}\alpha =0\Rightarrow \frac{{\mathrm{e}}^{-3\pi }}{2}\alpha =3\Rightarrow \alpha =6{\mathrm{e}}^{3\pi }\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{-3\mathrm{x}}({6\mathrm{e}}^{3\pi }\cos \left(\frac{\mathrm{x}}{2}\right)+{\mathrm{e}}^{3\pi }\sin \left(\frac{\mathrm{x}}{2}\right))$$$$={\mathrm{e}}^{3(\pi -\mathrm{x})}\{\cos \left(\frac{\mathrm{x}}{2}\right)+\sin \left(\frac{\mathrm{x}}{2}\right)\}$$

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