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Question Number 78878 by M±th+et£s last updated on 21/Jan/20

x^3 y′′′ − 3x^2 y′′+6xy′ −6y=x^4 ln(x),x>0

$${x}^{\mathrm{3}} \:{y}'''\:−\:\mathrm{3}{x}^{\mathrm{2}} {y}''+\mathrm{6}{xy}'\:−\mathrm{6}{y}={x}^{\mathrm{4}} \:{ln}\left({x}\right),{x}>\mathrm{0} \\ $$

Answered by mind is power last updated on 21/Jan/20

y=x^a Solution of homgenius⇒ x^3 y′′′−3x^2 y′′+6xy′−6y (a(a−1)(a−2)−3a(a−1)+6a−6)y(x)=0 ⇒ (a−1)(a(a−2)−3a+6)=0 (a−1)(a^2 −5a+6)=0 ⇒(a−1)(a−3)(a−2) ⇒a∈{1,2,3} y_1 =x,y_2 =x^2 ,y_3 =x^3 ,y_1 ,y_2 ,y_3 3 independent Solution ⇒ Y=ay_1 +by_2 +cy_3 Solution of Homgenius Equation let y=g(x)x particular Solution y′=xg′+g y′′=2g′+xg′′ y′′′=3g′′+xg′′′ ⇒x^3 (3g′′+xg′′′)−3x^2 (2g′+xg′′)+6x(xg′+g)−6xg(x)=x^4 ln(x) ⇒x^4 g′′′=x^4 ln(x)⇒g′′′(x)=ln(x) ⇒g′′=xln(x)−x g′(x)=((x^2 ln(x))/2)−(x^2 /4)−(x^2 /2)=((x^2 ln(x))/2)−((3x^2 )/4) g(x)=((x^3 ln(x))/6)−(x^3 /(18))−(x^3 /4)=((x^3 ln(x))/6)−((11x^3 )/(36)) y(x)=ax+bx^2 +cx^3 +(x^4 /6)(ln(x)−((11)/6))

$${y}={x}^{{a}} \:{Solution}\:{of}\:{homgenius}\Rightarrow \\ $$ $${x}^{\mathrm{3}} {y}'''−\mathrm{3}{x}^{\mathrm{2}} {y}''+\mathrm{6}{xy}'−\mathrm{6}{y} \\ $$ $$\left({a}\left({a}−\mathrm{1}\right)\left({a}−\mathrm{2}\right)−\mathrm{3}{a}\left({a}−\mathrm{1}\right)+\mathrm{6}{a}−\mathrm{6}\right){y}\left({x}\right)=\mathrm{0} \\ $$ $$\Rightarrow \\ $$ $$\left({a}−\mathrm{1}\right)\left({a}\left({a}−\mathrm{2}\right)−\mathrm{3}{a}+\mathrm{6}\right)=\mathrm{0} \\ $$ $$\left({a}−\mathrm{1}\right)\left({a}^{\mathrm{2}} −\mathrm{5}{a}+\mathrm{6}\right)=\mathrm{0} \\ $$ $$\Rightarrow\left({a}−\mathrm{1}\right)\left({a}−\mathrm{3}\right)\left({a}−\mathrm{2}\right) \\ $$ $$\Rightarrow{a}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$ $${y}_{\mathrm{1}} ={x},{y}_{\mathrm{2}} ={x}^{\mathrm{2}} ,{y}_{\mathrm{3}} ={x}^{\mathrm{3}} \:,{y}_{\mathrm{1}} ,{y}_{\mathrm{2}} ,{y}_{\mathrm{3}} \mathrm{3}\:{independent}\:{Solution}\:\Rightarrow \\ $$ $${Y}={ay}_{\mathrm{1}} +{by}_{\mathrm{2}} +{cy}_{\mathrm{3}} \:\:{Solution}\:{of}\:{Homgenius}\:{Equation} \\ $$ $${let}\:{y}={g}\left({x}\right){x}\:{particular}\:{Solution} \\ $$ $${y}'={xg}'+{g} \\ $$ $${y}''=\mathrm{2}{g}'+{xg}'' \\ $$ $${y}'''=\mathrm{3}{g}''+{xg}''' \\ $$ $$\Rightarrow{x}^{\mathrm{3}} \left(\mathrm{3}{g}''+{xg}'''\right)−\mathrm{3}{x}^{\mathrm{2}} \left(\mathrm{2}{g}'+{xg}''\right)+\mathrm{6}{x}\left({xg}'+{g}\right)−\mathrm{6}{xg}\left({x}\right)={x}^{\mathrm{4}} {ln}\left({x}\right) \\ $$ $$\Rightarrow{x}^{\mathrm{4}} {g}'''={x}^{\mathrm{4}} {ln}\left({x}\right)\Rightarrow{g}'''\left({x}\right)={ln}\left({x}\right) \\ $$ $$\Rightarrow{g}''={xln}\left({x}\right)−{x} \\ $$ $${g}'\left({x}\right)=\frac{{x}^{\mathrm{2}} {ln}\left({x}\right)}{\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}=\frac{{x}^{\mathrm{2}} {ln}\left({x}\right)}{\mathrm{2}}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$ $${g}\left({x}\right)=\frac{{x}^{\mathrm{3}} {ln}\left({x}\right)}{\mathrm{6}}−\frac{{x}^{\mathrm{3}} }{\mathrm{18}}−\frac{{x}^{\mathrm{3}} }{\mathrm{4}}=\frac{{x}^{\mathrm{3}} {ln}\left({x}\right)}{\mathrm{6}}−\frac{\mathrm{11}{x}^{\mathrm{3}} }{\mathrm{36}} \\ $$ $${y}\left({x}\right)={ax}+{bx}^{\mathrm{2}} +{cx}^{\mathrm{3}} +\frac{{x}^{\mathrm{4}} }{\mathrm{6}}\left({ln}\left({x}\right)−\frac{\mathrm{11}}{\mathrm{6}}\right) \\ $$ $$ \\ $$

Commented byM±th+et£s last updated on 21/Jan/20

god bless you sir

$${god}\:{bless}\:{you}\:{sir} \\ $$ $$ \\ $$

Commented bymind is power last updated on 21/Jan/20

Thanx you To sir

$${Thanx}\:{you}\:{To}\:{sir} \\ $$

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