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Question Number 173733 by Tawa11 last updated on 17/Jul/22

$$\mathrm{Show}\mathrm{that}{\mathrm{x}}^3+{\mathrm{y}}^3=1\mathrm{is}\mathrm{a}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{differential}$$$$\mathrm{equation}{20\mathrm{x}}^3+{3\mathrm{y}}^2\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+6\mathrm{y}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=0$$

Commented by kaivan.ahmadi last updated on 17/Jul/22

$$y^3=1-x^3\Rightarrow 3y^2{y}^{\prime }=-3x^2\Rightarrow y^2{y}^{\prime }=-x^2$$$$\Rightarrow y^{\prime }=-\frac{x^2}{y^2}$$$$ontheotherhandsince{y}^2{y}^{\prime }=-x^{2}so$$$$2{yy}^{\prime }.y^{\prime }+y^2{y}^{″}=-2x\Rightarrow y^{″}=-\frac{2x+2{yy}^{\prime 2}}{y^2}=$$$$-\frac{2x+2y\left(\frac{x^4}{y^4}\right)}{y^2}=-2\frac{{xy}^3+x^4}{y^5}$$$$now20x^3+3y^2{y}^{″}+6{yy}^{\prime 2}=$$$$20x^3+3y^2(-2\frac{{xy}^3+x^4}{y^5})+6y\left(\frac{x^4}{y^4}\right)=$$$$20x^3-6\frac{{xy}^3+x^4}{y^3}+6\frac{x^4}{y^3}=$$$$\frac{20x^3{y}^3-6{xy}^3-6x^4+6x^4}{y^3}=\frac{2y^{3}x(10x^2-3)}{y^3}$$$$=20x^3-6x$$$$$$

Commented by Tawa11 last updated on 17/Jul/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa11 last updated on 17/Jul/22

$$\mathrm{In}\mathrm{conclusion},\mathrm{that}\mathrm{means}{\mathrm{x}}^3+{\mathrm{y}}^3=1\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{solution}.$$

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