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Question Number 114980 by srijachakraborty last updated on 22/Sep/20

Answered by Dwaipayan Shikari last updated on 22/Sep/20

$$(y^4-2x^{3}y)+(x^4-2{xy}^3)\frac{dy}{dx}=0$$$$\frac{dy}{dx}=\frac{2x^{3}vx-v^4{x}^4}{x^4-2v^3{x}^4}(y=vx$$$$v+x\frac{dv}{dx}=\frac{2v-v^4}{1-2v^3}$$$$\frac{dv}{dx}=\frac{2v-v^4}{1-2v^3}-v$$$$\frac{dv}{dx}=\frac{v+v^4}{1-2v^3}$$$$\int \frac{(1-2v^3)dv}{v+v^4}=\int dx$$$$\int \frac{1+4v^3}{v+v^4}-\frac{6v^2}{(1+v^3)}=x+C$$$$log(v+v^4)-2log(1+v^3)=x+C$$$$log\left(\frac{v+v^4}{{(1+v^3)}^2}\right)=x+C$$$$\frac{v+v^4}{{(1+v^3)}^2}={Ce}^x$$$$\frac{\frac{y}{x}+{\left(\frac{y}{x}\right)}^4}{(x^3+y^3)}.x^6={Ce}^x$$$$\frac{x^{3}y+y^4}{(x^3+y^3)}.x^2={Ce}^x$$$$\frac{y+y^4}{{(1+y^3)}^2}=Ce$$$$\frac{y}{1+y^3}=Ce$$$$\frac{1}{2}=Ce$$$$C=\frac{1}{2e}$$$$\frac{x^{5}y+x^2{y}^4}{{(x^3+y^3)}^2}=\frac{1}{2}{e}^{x-1}$$

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