Question Number 116055 by Study last updated on 30/Sep/20
$$\mathrm{3}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{4}\frac{{dy}}{{dx}}+\mathrm{5}{y}=\mathrm{0}\:\:\:\:\:\:{y}=? \\ $$
Commented by mohammad17 last updated on 30/Sep/20
$$\left(\mathrm{3}{D}^{\mathrm{2}} +\mathrm{4}{D}+\mathrm{5}\right){y}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{3}{m}^{\mathrm{2}} +\mathrm{4}{m}+\mathrm{5}\right)=\mathrm{0} \\ $$$$ \\ $$$${m}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{60}}}{\mathrm{6}}\Rightarrow{m}_{\mathrm{1}} =−\frac{\mathrm{2}}{\mathrm{3}}−\frac{\sqrt{\mathrm{44}}}{\mathrm{6}}{i}\:\:{and}\:{m}_{\mathrm{2}} =−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\sqrt{\mathrm{44}}}{\mathrm{6}}{i} \\ $$$$ \\ $$$${y}={e}^{−\frac{\mathrm{2}}{\mathrm{3}}{x}} \left\{{C}_{\mathrm{1}} {cos}\left(\frac{\sqrt{\mathrm{44}}}{\mathrm{6}}{x}\right)+{C}_{\mathrm{2}} {sin}\left(\frac{\sqrt{\mathrm{44}}}{\mathrm{6}}{x}\right)\right\} \\ $$$$ \\ $$$$\ll{m}.{o}\gg \\ $$
Answered by Bird last updated on 30/Sep/20
$$\mathrm{3}{y}^{''} \:+\mathrm{4}{y}^{'} \:+\mathrm{5}=\mathrm{0}\:\Rightarrow\mathrm{3}{r}^{\mathrm{2}} \:+\mathrm{4}{r}\:\:+\mathrm{5}=\mathrm{0} \\ $$$$\Delta^{'\:} =\mathrm{4}−\mathrm{15}\:=−\mathrm{11}\:\Rightarrow \\ $$$${r}_{\mathrm{1}} =\frac{−\mathrm{2}+{i}\sqrt{\mathrm{11}}}{\mathrm{3}}\:{and}\:{r}_{\mathrm{2}} =\frac{−\mathrm{2}−{i}\sqrt{\mathrm{11}}}{\mathrm{3}} \\ $$$$\Rightarrow{y}\:=\alpha\:{e}^{{r}_{\mathrm{1}} {x}} +\beta{e}^{{r}_{\mathrm{2}} {x}} ={e}^{\frac{−\mathrm{2}{x}}{\mathrm{3}}} \left\{{acos}\left(\frac{\sqrt{\mathrm{11}}}{\mathrm{3}}{x}\right)+{bsin}\left(\frac{\sqrt{\mathrm{11}}}{\mathrm{3}}{x}\right)\right\} \\ $$$$ \\ $$