Question Number 108094 by Dwaipayan Shikari last updated on 14/Aug/20
$$\frac{{d}^{\mathrm{2}} \Psi}{{dt}^{\mathrm{2}} }+\frac{{d}\Psi}{{dt}}+\Psi=\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 14/Aug/20
$$\mathrm{y}^{''} \:+\mathrm{y}^{'} \:+\mathrm{y}\:=\mathrm{0}\:\rightarrow\mathrm{r}^{\mathrm{2}} +\mathrm{r}\:+\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}\:=−\mathrm{3}\:\Rightarrow\mathrm{r}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{r}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{e}^{−\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\mathrm{y}\:=\mathrm{a}\:\mathrm{e}^{\mathrm{r}_{\mathrm{1}} \mathrm{x}} \:+\mathrm{be}^{\mathrm{r}_{\mathrm{2}} \mathrm{x}} \:=\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{2}}} \left\{\alpha\:\mathrm{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{x}\right)+\beta\mathrm{sin}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{x}\right)\right\} \\ $$
Commented by Dwaipayan Shikari last updated on 14/Aug/20
Thanking you ��
Commented by mathmax by abdo last updated on 14/Aug/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$