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Question Number 91010 by jagoll last updated on 27/Apr/20
solve the diff eq y′′′−y′′+4y′−4y= e^x
$${solve}\:{the}\:{diff}\:{eq}\: \\ $$$${y}'''−{y}''+\mathrm{4}{y}'−\mathrm{4}{y}=\:{e}^{{x}} \\ $$
Commented by john santu last updated on 27/Apr/20
the characteristic equation w^3 −w^2 +4w −4 = 0 with roots are w = 1, ± 2i complementary solution y_c = K_1 e^x +K_2 cos 2x+K_3 sin 2x particular solution y_p = Qxe^x +Re^x y′= Qxe^x +(Q+R)e^x y′′= Qxe^x +(2Q+R)e^x y′′′ = Qxe^x +(3Q+R)e^x comparing coeff y′′′−y′′+4y′−4y = 5Qe^x = e^x Q = (1/5) general solution y = (1/5)xe^x +K_1 e^x +K_2 cos 2x+K_3 sin 2x
$${the}\:{characteristic}\:{equation}\: \\ $$$${w}^{\mathrm{3}} −{w}^{\mathrm{2}} +\mathrm{4}{w}\:−\mathrm{4}\:=\:\mathrm{0} \\ $$$${with}\:{roots}\:{are}\:{w}\:=\:\mathrm{1},\:\pm\:\mathrm{2}{i} \\ $$$${complementary}\:{solution} \\ $$$${y}_{{c}} \:=\:{K}_{\mathrm{1}} {e}^{{x}} +{K}_{\mathrm{2}} \mathrm{cos}\:\mathrm{2}{x}+{K}_{\mathrm{3}} \mathrm{sin}\:\mathrm{2}{x} \\ $$$${particular}\:{solution} \\ $$$${y}_{{p}} \:=\:{Qxe}^{{x}} +{Re}^{{x}} \\ $$$${y}'=\:{Qxe}^{{x}} +\left({Q}+{R}\right){e}^{{x}} \\ $$$${y}''=\:{Qxe}^{{x}} +\left(\mathrm{2}{Q}+{R}\right){e}^{{x}} \\ $$$${y}'''\:=\:{Qxe}^{{x}} +\left(\mathrm{3}{Q}+{R}\right){e}^{{x}} \\ $$$${comparing}\:{coeff} \\ $$$${y}'''−{y}''+\mathrm{4}{y}'−\mathrm{4}{y}\:=\:\mathrm{5}{Qe}^{{x}} =\:{e}^{{x}} \\ $$$${Q}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${general}\:{solution}\: \\ $$$${y}\:=\:\frac{\mathrm{1}}{\mathrm{5}}{xe}^{{x}} +{K}_{\mathrm{1}} {e}^{{x}} +{K}_{\mathrm{2}} \mathrm{cos}\:\mathrm{2}{x}+{K}_{\mathrm{3}} \mathrm{sin}\:\mathrm{2}{x} \\ $$
Commented by jagoll last updated on 27/Apr/20
thank you both
$${thank}\:{you}\:{both} \\ $$
Answered by MWSuSon last updated on 27/Apr/20
Auxillary equation m^3 −4m^2 +4m−4=0 (m−1)(m^2 +4)=0 m=1,±i2 y_c =C_1 e^x +C_2 cos (2x)+C_3 sin (2x) y_p =(1/(D^3 −D^2 +4D−4))e^x =(1/(1−1+4−4))e^x which is undefinded so y_p =x(1/(3D^2 −2D+4))e^x y_p =x(1/(3−2+4))e^x =((xe^x )/5) y=C_1 e^x +C_2 cos (2x)+C_3 sin (2x)+((xe^x )/5)
$${Auxillary}\:{equation} \\ $$$${m}^{\mathrm{3}} −\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}{m}−\mathrm{4}=\mathrm{0} \\ $$$$\left({m}−\mathrm{1}\right)\left({m}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{0} \\ $$$${m}=\mathrm{1},\pm{i}\mathrm{2} \\ $$$${y}_{{c}} ={C}_{\mathrm{1}} {e}^{{x}} +{C}_{\mathrm{2}} \mathrm{cos}\:\left(\mathrm{2}{x}\right)+{C}_{\mathrm{3}} \mathrm{sin}\:\left(\mathrm{2}{x}\right) \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{3}} −{D}^{\mathrm{2}} +\mathrm{4}{D}−\mathrm{4}}{e}^{{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{1}+\mathrm{4}−\mathrm{4}}{e}^{{x}} \:{which}\:{is}\:{undefinded} \\ $$$${so}\:{y}_{{p}} ={x}\frac{\mathrm{1}}{\mathrm{3}{D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{4}}{e}^{{x}} \\ $$$${y}_{{p}} ={x}\frac{\mathrm{1}}{\mathrm{3}−\mathrm{2}+\mathrm{4}}{e}^{{x}} =\frac{{xe}^{{x}} }{\mathrm{5}} \\ $$$${y}={C}_{\mathrm{1}} {e}^{{x}} +{C}_{\mathrm{2}} \mathrm{cos}\:\left(\mathrm{2}{x}\right)+{C}_{\mathrm{3}} \mathrm{sin}\:\left(\mathrm{2}{x}\right)+\frac{{xe}^{{x}} }{\mathrm{5}} \\ $$

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