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Question-48920




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^(−2mx) sin4mx  (dy/dx)=e^(−2mx) ×4mcos4mx−2me^(−2mx) sin4mx  (dy/dx)=4me^(−2mx) cos4mx−2my  (d^2 y/dx^2 )=4me^(−2mx) ×−4msin4mx−4m×2me^(−2mx) cos4mx−2m(dy/dx)  (d^2 y/dx^2 )=−16m^2 y−2m((dy/dx)+2my)−2m(dy/dx)  (d^2 y/dx^2 )=−16m^2 y−4m(dy/dx)−4m^2 y  (d^2 y/dx^2 )+20m^2 y+4m(dy/dx)=0
$${y}={e}^{−\mathrm{2}{mx}} {sin}\mathrm{4}{mx} \\ $$$$\frac{{dy}}{{dx}}={e}^{−\mathrm{2}{mx}} ×\mathrm{4}{mcos}\mathrm{4}{mx}−\mathrm{2}{me}^{−\mathrm{2}{mx}} {sin}\mathrm{4}{mx} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}{me}^{−\mathrm{2}{mx}} {cos}\mathrm{4}{mx}−\mathrm{2}{my} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{4}{me}^{−\mathrm{2}{mx}} ×−\mathrm{4}{msin}\mathrm{4}{mx}−\mathrm{4}{m}×\mathrm{2}{me}^{−\mathrm{2}{mx}} {cos}\mathrm{4}{mx}−\mathrm{2}{m}\frac{{dy}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−\mathrm{16}{m}^{\mathrm{2}} {y}−\mathrm{2}{m}\left(\frac{{dy}}{{dx}}+\mathrm{2}{my}\right)−\mathrm{2}{m}\frac{{dy}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=−\mathrm{16}{m}^{\mathrm{2}} {y}−\mathrm{4}{m}\frac{{dy}}{{dx}}−\mathrm{4}{m}^{\mathrm{2}} {y} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{20}{m}^{\mathrm{2}} {y}+\mathrm{4}{m}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$

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