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Question Number 46382 by peter frank last updated on 24/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Oct/18

$$\frac{dy}{dx}=-\frac{y^2+y+1}{x^2+x+1}$$$$\frac{dy}{y^2+y+1}+\frac{dx}{x^2+x+1}=0$$$$\int \frac{dy}{y^2+y+1}+\int \frac{dx}{x^2+x+1}=C$$$$\int \frac{dy}{y^2+2\times y\times \frac{1}{2}+\frac{1}{4}+\frac{3}{4}}+\int \frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$\frac{2}{\sqrt{3}}{tan}^{-1}\left(\frac{y+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+\frac{2}{\sqrt{3}}{tan}^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)=c$$$${tan}^{-1}\left(\frac{2y+1}{\sqrt{3}}\right)+{tan}^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)=\frac{\sqrt{3}c}{2}$$$${tan}^{-1}\left(\frac{\frac{2y+1}{\sqrt{3}}+\frac{2x+1}{\sqrt{3}}}{1-\left(\frac{4xy+2y+2x+1}{3}\right)}\right)=C_1$$$$\frac{(2x+2y+2)}{\sqrt{3}\left(\frac{3-4xy-2x-2y-1}{3}\right)}={tanC}_1$$$$\frac{\sqrt{3}\times 2(x+y+1)}{2(1-2xy-x-y)}={tanC}_1$$$$\frac{x+y+1}{1-2xy-x-y}=\frac{{tanC}_1}{\sqrt{3}}=A$$$$(x+y+1)=A(1-2xy-x-y)$$$$$$$$$$

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