Question Number 123 by novrya last updated on 25/Jan/15
$${Solve}\:{the}\:{differential}\:{equation}: \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{2}\frac{{dy}}{{dx}}+{y}={e}^{\mathrm{4}{x}} \\ $$
Answered by sudhanshur last updated on 07/Dec/14
$$\mathrm{Let}\:{y}={Ae}^{\mathrm{4}{x}} \: \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}{Ae}^{\mathrm{4}{x}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{16}{Ae}^{\mathrm{4}{x}} \\ $$$$\mathrm{16}{Ae}^{\mathrm{4}{x}} −\mathrm{8}{Ae}^{\mathrm{4}{x}} +{Ae}^{\mathrm{4}{x}} ={e}^{\mathrm{4}{x}} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{9}}{e}^{\mathrm{4}{x}} \\ $$
Commented by 123456 last updated on 14/Dec/14
$${y}''−\mathrm{2}{y}'+\mathrm{1}=\mathrm{0} \\ $$$$\lambda^{\mathrm{2}} −\mathrm{2}\lambda+\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right) \\ $$$$\:\:\:\:=\mathrm{4}−\mathrm{4}=\mathrm{0} \\ $$$$\lambda=\frac{−\left(−\mathrm{2}\right)+\sqrt{\mathrm{0}}}{\mathrm{2}\left(\mathrm{1}\right)}=\mathrm{1}\: \\ $$$${y}_{{n}} =\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} {x}\right){e}^{{x}} \\ $$$${y}={y}_{{n}} +{y}_{{p}} =\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} {x}\right){e}^{{x}} +\frac{\mathrm{1}}{\mathrm{9}}{e}^{\mathrm{4}{x}} \\ $$
Commented by sudhanshur last updated on 15/Dec/14
$$\boldsymbol{\mathrm{Thank}}\:\boldsymbol{\mathrm{You}},\:\mathrm{123456}! \\ $$