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Question Number 123 by novrya last updated on 25/Jan/15
Solve the differential equation: (d^2 y/dx^2 )−2(dy/dx)+y=e^(4x)
$${Solve}\:{the}\:{differential}\:{equation}: \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{2}\frac{{dy}}{{dx}}+{y}={e}^{\mathrm{4}{x}} \\ $$
Answered by sudhanshur last updated on 07/Dec/14
Let y=Ae^(4x) (dy/dx)=4Ae^(4x) (d^2 y/dx^2 )=16Ae^(4x) 16Ae^(4x) −8Ae^(4x) +Ae^(4x) =e^(4x) A=(1/9) y=(1/9)e^(4x)
$$\mathrm{Let}\:{y}={Ae}^{\mathrm{4}{x}} \: \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}{Ae}^{\mathrm{4}{x}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{16}{Ae}^{\mathrm{4}{x}} \\ $$$$\mathrm{16}{Ae}^{\mathrm{4}{x}} −\mathrm{8}{Ae}^{\mathrm{4}{x}} +{Ae}^{\mathrm{4}{x}} ={e}^{\mathrm{4}{x}} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{9}}{e}^{\mathrm{4}{x}} \\ $$
Commented by 123456 last updated on 14/Dec/14
y′′−2y′+1=0 λ^2 −2λ+1=0 Δ=(−2)^2 −4(1)(1) =4−4=0 λ=((−(−2)+(√0))/(2(1)))=1 y_n =(c_1 +c_2 x)e^x y=y_n +y_p =(c_1 +c_2 x)e^x +(1/9)e^(4x)
$${y}''−\mathrm{2}{y}'+\mathrm{1}=\mathrm{0} \\ $$$$\lambda^{\mathrm{2}} −\mathrm{2}\lambda+\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right) \\ $$$$\:\:\:\:=\mathrm{4}−\mathrm{4}=\mathrm{0} \\ $$$$\lambda=\frac{−\left(−\mathrm{2}\right)+\sqrt{\mathrm{0}}}{\mathrm{2}\left(\mathrm{1}\right)}=\mathrm{1}\: \\ $$$${y}_{{n}} =\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} {x}\right){e}^{{x}} \\ $$$${y}={y}_{{n}} +{y}_{{p}} =\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} {x}\right){e}^{{x}} +\frac{\mathrm{1}}{\mathrm{9}}{e}^{\mathrm{4}{x}} \\ $$
Commented by sudhanshur last updated on 15/Dec/14
Thank You, 123456!
$$\boldsymbol{\mathrm{Thank}}\:\boldsymbol{\mathrm{You}},\:\mathrm{123456}! \\ $$

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