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Question Number 126223 by liberty last updated on 18/Dec/20

$$\frac{dy}{dx}+\frac{x^{2}y}{x^3+y^3}=0$$

Commented by bramlexs22 last updated on 18/Dec/20

$$v+x\frac{dv}{dx}=\frac{x^{3}v}{x^3(1+v^3)};[v=\frac{y}{x}]$$$$v+x\frac{dv}{dx}=\frac{v}{1+v^3}$$$$x\frac{dv}{dx}=\frac{-v^4}{1+v^3};\frac{(1+v^3)dv}{v^4}+\frac{dx}{x}=0$$$$\int (v^{-4}+v^{-1})dv+\int \frac{dx}{x}=C$$$$-\frac{1}{3v^3}+\ln \left(vx\right)=C$$$$-\frac{x^3}{3y^3}+\ln \left(y\right)=C$$$$\frac{x^3}{3y^3}=\ln \left(\lambda y\right)\Rightarrow \frac{3y^3}{x^3}=\frac{1}{\ln \left(\lambda y\right)}$$$$y^3=\frac{x^3}{3\ln \left(\lambda y\right)}.$$$$$$

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