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Question Number 173733 by Tawa11 last updated on 17/Jul/22
Show that x^3 + y^3 = 1 is a solution to the differential equation 20x^3 + 3y^2 (d^2 y/dx^2 ) + 6y((dy/dx))^2 = 0
$$\mathrm{Show}\:\mathrm{that}\:\:\:\:\mathrm{x}^{\mathrm{3}} \:\:+\:\:\mathrm{y}^{\mathrm{3}} \:\:=\:\:\mathrm{1}\:\:\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation}\:\:\:\:\mathrm{20x}^{\mathrm{3}} \:\:\:+\:\:\:\mathrm{3y}^{\mathrm{2}} \:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:\:\:\:+\:\:\:\mathrm{6y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \:\:\:=\:\:\:\:\mathrm{0} \\ $$
Commented by kaivan.ahmadi last updated on 17/Jul/22
y^3 =1−x^3 ⇒3y^2 y′=−3x^2 ⇒y^2 y′=−x^2 ⇒y′=−(x^2 /y^2 ) on the other hand since y^2 y′=−x^(2 ) so 2yy′.y′+y^2 y′′=−2x⇒y′′=−((2x+2yy′^2 )/y^2 )= −((2x+2y((x^4 /y^4 )))/y^2 )=−2 ((xy^3 +x^4 )/y^5 ) now 20x^3 +3y^2 y′′+6yy′^2 = 20x^3 +3y^2 (−2((xy^3 +x^4 )/y^5 ))+6y((x^4 /y^4 ))= 20x^3 −6 ((xy^3 +x^4 )/y^3 )+6(x^4 /y^3 )= ((20x^3 y^3 −6xy^3 −6x^4 +6x^4 )/y^3 )=((2y^3 x(10x^2 −3))/y^3 ) =20x^3 −6x
$${y}^{\mathrm{3}} =\mathrm{1}−{x}^{\mathrm{3}} \Rightarrow\mathrm{3}{y}^{\mathrm{2}} {y}'=−\mathrm{3}{x}^{\mathrm{2}} \Rightarrow{y}^{\mathrm{2}} {y}'=−{x}^{\mathrm{2}} \\ $$$$\Rightarrow{y}'=−\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} } \\ $$$${on}\:{the}\:{other}\:{hand}\:{since}\:{y}^{\mathrm{2}} {y}'=−{x}^{\mathrm{2}\:} {so} \\ $$$$\mathrm{2}{yy}'.{y}'+{y}^{\mathrm{2}} {y}''=−\mathrm{2}{x}\Rightarrow{y}''=−\frac{\mathrm{2}{x}+\mathrm{2}{yy}'^{\mathrm{2}} }{{y}^{\mathrm{2}} }= \\ $$$$−\frac{\mathrm{2}{x}+\mathrm{2}{y}\left(\frac{{x}^{\mathrm{4}} }{{y}^{\mathrm{4}} }\right)}{{y}^{\mathrm{2}} }=−\mathrm{2}\:\frac{{xy}^{\mathrm{3}} +{x}^{\mathrm{4}} }{{y}^{\mathrm{5}} } \\ $$$${now}\:\mathrm{20}{x}^{\mathrm{3}} +\mathrm{3}{y}^{\mathrm{2}} {y}''+\mathrm{6}{yy}'^{\mathrm{2}} = \\ $$$$\:\mathrm{20}{x}^{\mathrm{3}} +\mathrm{3}{y}^{\mathrm{2}} \left(−\mathrm{2}\frac{{xy}^{\mathrm{3}} +{x}^{\mathrm{4}} }{{y}^{\mathrm{5}} }\right)+\mathrm{6}{y}\left(\frac{{x}^{\mathrm{4}} }{{y}^{\mathrm{4}} }\right)= \\ $$$$\mathrm{20}{x}^{\mathrm{3}} −\mathrm{6}\:\frac{{xy}^{\mathrm{3}} +{x}^{\mathrm{4}} }{{y}^{\mathrm{3}} }+\mathrm{6}\frac{{x}^{\mathrm{4}} }{{y}^{\mathrm{3}} }= \\ $$$$\frac{\mathrm{20}{x}^{\mathrm{3}} {y}^{\mathrm{3}} −\mathrm{6}{xy}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{4}} }{{y}^{\mathrm{3}} }=\frac{\mathrm{2}{y}^{\mathrm{3}} {x}\left(\mathrm{10}{x}^{\mathrm{2}} −\mathrm{3}\right)}{{y}^{\mathrm{3}} } \\ $$$$=\mathrm{20}{x}^{\mathrm{3}} −\mathrm{6}{x} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 17/Jul/22
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 17/Jul/22
In conclusion, that means x^3 + y^3 = 1 is not a solution.
$$\mathrm{In}\:\mathrm{conclusion},\:\mathrm{that}\:\mathrm{means}\:\:\mathrm{x}^{\mathrm{3}} \:\:+\:\:\mathrm{y}^{\mathrm{3}} \:\:=\:\:\mathrm{1}\:\:\mathrm{is}\:\:\mathrm{not}\:\mathrm{a}\:\mathrm{solution}. \\ $$

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