Question Number 48920 by peter frank last updated on 30/Nov/18
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Nov/18
$${y}={e}^{−\mathrm{2}{mx}} {sin}\mathrm{4}{mx} \\ $$$$\frac{{dy}}{{dx}}={e}^{−\mathrm{2}{mx}} ×\mathrm{4}{mcos}\mathrm{4}{mx}−\mathrm{2}{me}^{−\mathrm{2}{mx}} {sin}\mathrm{4}{mx} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}{me}^{−\mathrm{2}{mx}} {cos}\mathrm{4}{mx}−\mathrm{2}{my} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{4}{me}^{−\mathrm{2}{mx}} ×−\mathrm{4}{msin}\mathrm{4}{mx}−\mathrm{4}{m}×\mathrm{2}{me}^{−\mathrm{2}{mx}} {cos}\mathrm{4}{mx}−\mathrm{2}{m}\frac{{dy}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−\mathrm{16}{m}^{\mathrm{2}} {y}−\mathrm{2}{m}\left(\frac{{dy}}{{dx}}+\mathrm{2}{my}\right)−\mathrm{2}{m}\frac{{dy}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−\mathrm{16}{m}^{\mathrm{2}} {y}−\mathrm{4}{m}\frac{{dy}}{{dx}}−\mathrm{4}{m}^{\mathrm{2}} {y} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{20}{m}^{\mathrm{2}} {y}+\mathrm{4}{m}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$