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Question Number 108094 by Dwaipayan Shikari last updated on 14/Aug/20

$$\frac{d^2\mathrm{\Psi }}{{dt}^2}+\frac{d\mathrm{\Psi }}{dt}+\mathrm{\Psi }=0$$

Answered by mathmax by abdo last updated on 14/Aug/20

$${\mathrm{y}}^{{}^{″}}+{\mathrm{y}}^{{}^{\prime }}+\mathrm{y}=0\to {\mathrm{r}}^2+\mathrm{r}+1=0$$$$\mathrm{\Delta }=1-4=-3\Rightarrow {\mathrm{r}}_1=\frac{-1+\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{\mathrm{i}\frac{2\pi }{3}}\mathrm{and}{\mathrm{r}}_2=\frac{-1-\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{-\mathrm{i}\frac{2\pi }{3}}$$$$\Rightarrow \mathrm{y}=\mathrm{a}{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{x}}+{\mathrm{be}}^{{\mathrm{r}}_2\mathrm{x}}={\mathrm{e}}^{-\frac{\mathrm{x}}{2}}\{\alpha \cos \left(\frac{\sqrt{3}}{2}\mathrm{x}\right)+\beta \sin \left(\frac{\sqrt{3}}{2}\mathrm{x}\right)\}$$

Commented by Dwaipayan Shikari last updated on 14/Aug/20

Thanking you ��

Commented by mathmax by abdo last updated on 14/Aug/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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