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solve-D-2-2D-1-y-e-x-ln-x-by-using-the-method-of-variation-of-parameters-

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Question Number 119065 by benjo_mathlover last updated on 21/Oct/20
solve (D^2 −2D+1)y = e^x ln x by using the method of variation of parameters.
$${solve}\:\left({D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{1}\right){y}\:=\:{e}^{{x}} \:\mathrm{ln}\:{x}\: \\ $$$${by}\:{using}\:{the}\:{method}\:{of}\:{variation} \\ $$$${of}\:{parameters}. \\ $$
Answered by Olaf last updated on 22/Oct/20
y = e^x u y′ = (u′+u)e^x y′′ = (u′′+2u′+u)e^x u′′+2u′+u−2(u′+u)+u = lnx u′′ = lnx u′ = ∫lnxdx = xlnx−x+C_1 u = ∫(xlnx−x+C_1 )dx u = (x^2 /2)lnx−∫(x^2 /2).(1/x)dx−(x^2 /2)+C_1 x+C_2 u = (x^2 /2)lnx−(x^2 /4)−(x^2 /2)+C_1 x+C_2 u = (x^2 /2)lnx−((3x^2 )/4)+C_1 x+C_2 y = ((x^2 /2)lnx−((3x^2 )/4)+C_1 x+C_2 )e^x
$${y}\:=\:{e}^{{x}} {u} \\ $$$${y}'\:=\:\left({u}'+{u}\right){e}^{{x}} \\ $$$${y}''\:=\:\left({u}''+\mathrm{2}{u}'+{u}\right){e}^{{x}} \\ $$$$ \\ $$$${u}''+\mathrm{2}{u}'+{u}−\mathrm{2}\left({u}'+{u}\right)+{u}\:=\:\mathrm{ln}{x} \\ $$$${u}''\:=\:\mathrm{ln}{x} \\ $$$${u}'\:=\:\int\mathrm{ln}{xdx}\:=\:{x}\mathrm{ln}{x}−{x}+\mathrm{C}_{\mathrm{1}} \\ $$$${u}\:=\:\int\left({x}\mathrm{ln}{x}−{x}+\mathrm{C}_{\mathrm{1}} \right){dx} \\ $$$${u}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}{x}−\int\frac{{x}^{\mathrm{2}} }{\mathrm{2}}.\frac{\mathrm{1}}{{x}}{dx}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{C}_{\mathrm{1}} {x}+\mathrm{C}_{\mathrm{2}} \\ $$$${u}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{C}_{\mathrm{1}} {x}+\mathrm{C}_{\mathrm{2}} \\ $$$${u}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}{x}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{C}_{\mathrm{1}} {x}+\mathrm{C}_{\mathrm{2}} \\ $$$${y}\:=\:\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}{x}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{C}_{\mathrm{1}} {x}+\mathrm{C}_{\mathrm{2}} \right){e}^{{x}} \\ $$
Commented by benjo_mathlover last updated on 22/Oct/20
thank you
$${thank}\:{you} \\ $$
Answered by john santu last updated on 27/Oct/20
For the homogenous solution we solve the characteristic equation λ^2 −2λ+1 = 0 , which has a double root λ=1 hence the homogenous solution can be written as y_h = (C_1 +C_2 x)e^x using the homogenous solution we can now begin using variation of parameters method let y_p = u.e^x + v.xe^x Differentiating yields y ′ = (ue^x +u′e^x )+(v.(x+1)e^x +v′.xe^x ) set u′e^x + v′xe^x = 0 or u′+xv′ = 0 we get y′ = ue^x +v.(x+1)e^x Differentiating again gives y′′ = ue^x +u′e^x +v(x+2)e^x +v′(x+1)e^x substituting into the original differential equation (ue^x +u′e^x +v(x+2)e^x +v′(x+1)e^x )− 2(ue^x +v(x+1)e^x ) + (ue^x +v.xe^x ) = e^x .ln x this reduces to u′+(x+1)v′ = ln x now we have two equation in u′ and v′ { ((u′+xv′ = 0)),((u′+(x+1)v′ = ln x)) :} (((1 x)),((1 x+1)) ) (((u′)),((v′)) ) = ((( 0)),((ln x)) ) → { ((u′ = −x ln x)),((v′ = ln x)) :} integrating with by parts gives { ((u = −(1/2)x^2 ln x + (1/4)x^(2 ) )),((v = x ln x−x )) :} thus particular solution equal to y_p = −(1/2)x^2 e^x ln x + (1/4)x^2 e^x + x^2 e^x ln x −x^2 e^x y_p = (1/2)x^2 e^x ln x −(3/4)x^2 e^x Therefore we conclude that a general solution y = y_h + y_p y = (C_1 +C_2 x)e^x + (1/4)(2ln x−3)x^2 e^x
$${For}\:{the}\:{homogenous}\:{solution}\:{we}\:{solve}\:{the}\: \\ $$$${characteristic}\:{equation} \\ $$$$\lambda^{\mathrm{2}} −\mathrm{2}\lambda+\mathrm{1}\:=\:\mathrm{0}\:,\:{which}\:{has}\:{a}\:{double}\:{root}\:\lambda=\mathrm{1} \\ $$$${hence}\:{the}\:{homogenous}\:{solution}\:{can}\:{be}\: \\ $$$${written}\:{as}\:{y}_{{h}} \:=\:\left({C}_{\mathrm{1}} +{C}_{\mathrm{2}} {x}\right){e}^{{x}} \\ $$$${using}\:{the}\:{homogenous}\:{solution}\:{we}\: \\ $$$${can}\:{now}\:{begin}\:{using}\:{variation}\:{of} \\ $$$${parameters}\:{method}\: \\ $$$${let}\:{y}_{{p}} \:=\:{u}.{e}^{{x}} \:+\:{v}.{xe}^{{x}} \\ $$$${Differentiating}\:{yields}\: \\ $$$${y}\:'\:=\:\left({ue}^{{x}} +{u}'{e}^{{x}} \right)+\left({v}.\left({x}+\mathrm{1}\right){e}^{{x}} +{v}'.{xe}^{{x}} \right) \\ $$$${set}\:{u}'{e}^{{x}} \:+\:{v}'{xe}^{{x}} \:=\:\mathrm{0}\:{or}\:{u}'+{xv}'\:=\:\mathrm{0} \\ $$$${we}\:{get}\:{y}'\:=\:{ue}^{{x}} +{v}.\left({x}+\mathrm{1}\right){e}^{{x}} \\ $$$${Differentiating}\:{again}\:{gives}\: \\ $$$${y}''\:=\:{ue}^{{x}} +{u}'{e}^{{x}} +{v}\left({x}+\mathrm{2}\right){e}^{{x}} +{v}'\left({x}+\mathrm{1}\right){e}^{{x}} \\ $$$${substituting}\:{into}\:{the}\:{original}\: \\ $$$${differential}\:{equation}\: \\ $$$$\left({ue}^{{x}} +{u}'{e}^{{x}} +{v}\left({x}+\mathrm{2}\right){e}^{{x}} +{v}'\left({x}+\mathrm{1}\right){e}^{{x}} \right)− \\ $$$$\mathrm{2}\left({ue}^{{x}} +{v}\left({x}+\mathrm{1}\right){e}^{{x}} \right)\:+\:\left({ue}^{{x}} +{v}.{xe}^{{x}} \right)\:=\:{e}^{{x}} .\mathrm{ln}\:{x} \\ $$$${this}\:{reduces}\:{to}\: \\ $$$${u}'+\left({x}+\mathrm{1}\right){v}'\:=\:\mathrm{ln}\:{x} \\ $$$${now}\:{we}\:{have}\:{two}\:{equation}\:{in}\:{u}'\:{and}\:{v}' \\ $$$$\begin{cases}{{u}'+{xv}'\:=\:\mathrm{0}}\\{{u}'+\left({x}+\mathrm{1}\right){v}'\:=\:\mathrm{ln}\:{x}}\end{cases} \\ $$$$\begin{pmatrix}{\mathrm{1}\:\:\:\:\:{x}}\\{\mathrm{1}\:\:\:{x}+\mathrm{1}}\end{pmatrix}\:\begin{pmatrix}{{u}'}\\{{v}'}\end{pmatrix}\:=\:\begin{pmatrix}{\:\:\:\mathrm{0}}\\{\mathrm{ln}\:{x}}\end{pmatrix} \\ $$$$\rightarrow\begin{cases}{{u}'\:=\:−{x}\:\mathrm{ln}\:{x}}\\{{v}'\:=\:\mathrm{ln}\:{x}}\end{cases} \\ $$$${integrating}\:{with}\:{by}\:{parts}\:{gives} \\ $$$$\begin{cases}{{u}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \mathrm{ln}\:{x}\:+\:\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}\:} }\\{{v}\:=\:{x}\:\mathrm{ln}\:{x}−{x}\:}\end{cases} \\ $$$${thus}\:{particular}\:{solution}\:{equal}\:{to} \\ $$$${y}_{{p}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:{e}^{{x}} \:\mathrm{ln}\:{x}\:+\:\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}} \:{e}^{{x}} \:\: \\ $$$$\:\:\:\:\:\:+\:{x}^{\mathrm{2}} \:{e}^{{x}} \:\mathrm{ln}\:{x}\:−{x}^{\mathrm{2}} {e}^{{x}} \: \\ $$$${y}_{{p}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:{e}^{{x}} \:\mathrm{ln}\:{x}\:−\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} {e}^{{x}} \\ $$$${Therefore}\:{we}\:{conclude}\:{that}\:{a}\:{general} \\ $$$${solution}\:{y}\:=\:{y}_{{h}} \:+\:{y}_{{p}} \\ $$$${y}\:=\:\left({C}_{\mathrm{1}} +{C}_{\mathrm{2}} {x}\right){e}^{{x}} +\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2ln}\:{x}−\mathrm{3}\right){x}^{\mathrm{2}} {e}^{{x}} \\ $$
Answered by TANMAY PANACEA last updated on 22/Oct/20
P.I=((e^x lnx)/((D−1)^2 ))=e^x ×(1/((D+1−1)^2 ))×lnx =(e^x /1)×((lnx)/D^2 ) (1/D)×lnx=xlnx−x (1/D^2 )lnx=∫(xlnx−x)dx =lnx×(x^2 /2)−∫(1/x)×(x^2 /2)dx−(x^2 /2) =((x^2 lnx)/2)−(x^2 /4)−(x^2 /2) =(x^2 /2)(lnx−(1/2)−1)=(x^2 /2)(((−3)/2)+lnx) snsser ((e^x ×x^2 )/2)o(((−3)/2)+lnx)
$${P}.{I}=\frac{{e}^{{x}} {lnx}}{\left({D}−\mathrm{1}\right)^{\mathrm{2}} }={e}^{{x}} ×\frac{\mathrm{1}}{\left({D}+\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} }×{lnx} \\ $$$$=\frac{{e}^{{x}} }{\mathrm{1}}×\frac{{lnx}}{{D}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{D}}×{lnx}={xlnx}−{x} \\ $$$$\frac{\mathrm{1}}{{D}^{\mathrm{2}} }{lnx}=\int\left({xlnx}−{x}\right){dx} \\ $$$$={lnx}×\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\int\frac{\mathrm{1}}{{x}}×\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{dx}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{{x}^{\mathrm{2}} {lnx}}{\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left({lnx}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{−\mathrm{3}}{\mathrm{2}}+{lnx}\right) \\ $$$${snsser} \\ $$$$\frac{{e}^{{x}} ×{x}^{\mathrm{2}} }{\mathrm{2}}{o}\left(\frac{−\mathrm{3}}{\mathrm{2}}+{lnx}\right) \\ $$

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