Question-46382

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

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

(dy/dx)=−((y^2 +y+1)/(x^2 +x+1)) (dy/(y^2 +y+1))+(dx/(x^2 +x+1))=0 ∫(dy/(y^2 +y+1))+∫(dx/(x^2 +x+1))=C ∫(dy/(y^2 +2×y×(1/2)+(1/4)+(3/4)))+∫(dx/((x+(1/2))^2 +(((√3)/2))^2 )) (2/( (√3)))tan^(−1) (((y+(1/2))/((√3)/2)))+(2/( (√3)))tan^(−1) (((x+(1/2))/((√3)/2)))=c tan^(−1) (((2y+1)/( (√3))))+tan^(−1) (((2x+1)/( (√3))))=(((√3) c)/2) tan^(−1) (((((2y+1)/( (√3)))+((2x+1)/( (√3))))/(1−(((4xy+2y+2x+1)/3)))))=C_1 (((2x+2y+2))/( (√3) (((3−4xy−2x−2y−1)/3))))=tanC_1 (((√3) ×2(x+y+1))/(2(1−2xy−x−y)))=tanC_1 ((x+y+1)/(1−2xy−x−y))=((tanC_1 )/( (√3)))=A (x+y+1)=A(1−2xy−x−y)

$$\frac{{dy}}{{dx}}=−\frac{{y}^{\mathrm{2}} +{y}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$$\frac{{dy}}{{y}^{\mathrm{2}} +{y}+\mathrm{1}}+\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}=\mathrm{0} \\ $$$$\int\frac{{dy}}{{y}^{\mathrm{2}} +{y}+\mathrm{1}}+\int\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}={C} \\ $$$$\int\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{2}×{y}×\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}+\int\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\frac{{y}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\frac{{x}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)={c} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{y}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\frac{\sqrt{\mathrm{3}}\:{c}}{\mathrm{2}} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{2}{y}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}−\left(\frac{\mathrm{4}{xy}+\mathrm{2}{y}+\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}}\right)}\right)={C}_{\mathrm{1}} \\ $$$$\frac{\left(\mathrm{2}{x}+\mathrm{2}{y}+\mathrm{2}\right)}{\:\sqrt{\mathrm{3}}\:\left(\frac{\mathrm{3}−\mathrm{4}{xy}−\mathrm{2}{x}−\mathrm{2}{y}−\mathrm{1}}{\mathrm{3}}\right)}={tanC}_{\mathrm{1}} \\ $$$$\frac{\sqrt{\mathrm{3}}\:×\mathrm{2}\left({x}+{y}+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{xy}−{x}−{y}\right)}={tanC}_{\mathrm{1}} \\ $$$$\frac{{x}+{y}+\mathrm{1}}{\mathrm{1}−\mathrm{2}{xy}−{x}−{y}}=\frac{{tanC}_{\mathrm{1}} }{\:\sqrt{\mathrm{3}}}={A} \\ $$$$\left({x}+{y}+\mathrm{1}\right)={A}\left(\mathrm{1}−\mathrm{2}{xy}−{x}−{y}\right) \\ $$$$ \\ $$$$ \\ $$

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