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Question Number 117228 by bemath last updated on 10/Oct/20

(((x+y)/(y−1))) dx −(1/2)(((x+1)/(y−1)))^2 dy = 0

$$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right)\:{dx}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+\mathrm{1}}{{y}−\mathrm{1}}\right)^{\mathrm{2}} {dy}\:=\:\mathrm{0} \\ $$

Answered by TANMAY PANACEA last updated on 10/Oct/20

(((x+y)/(y−1)))dx−(1/2)(((x+y)/(y−1))−1)^2 dy=0 (((x+y)/(y−1)))dx−(1/2)(((x+y)/(y−1)))^2 dy+(((x+y)/(y−1)))dy−(1/2)dy=0 (((x+y)/(y−1)))(dx+dy)−(1/2)(((x+y)/(y−1)))^2 dy−(dy/2)=0 (x+y)d(x+y)−(1/2)(x+y)^2 (y−1)d(y−1)−(dy/2)=0 (x+y)d(x+y)−(1/2)(x+y)d(y−1)^2 ×(1/2)−(dy/2)=0 d(x+y)^2 −(x+y)×(1/2)d(y−1)^2 −((d(y−1))/2)=0 2d(x+y)^2 −(x+y)d(y−1)^2 −d(y−1)=0 wait

$$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right){dx}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+{y}}{{y}−\mathrm{1}}−\mathrm{1}\right)^{\mathrm{2}} {dy}=\mathrm{0} \\ $$$$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right){dx}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right)^{\mathrm{2}} {dy}+\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right){dy}−\frac{\mathrm{1}}{\mathrm{2}}{dy}=\mathrm{0} \\ $$$$\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right)\left({dx}+{dy}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{x}+{y}}{{y}−\mathrm{1}}\right)^{\mathrm{2}} {dy}−\frac{{dy}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left({x}+{y}\right){d}\left({x}+{y}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}\right)^{\mathrm{2}} \left({y}−\mathrm{1}\right){d}\left({y}−\mathrm{1}\right)−\frac{{dy}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left({x}+{y}\right){d}\left({x}+{y}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}\right){d}\left({y}−\mathrm{1}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{2}}−\frac{{dy}}{\mathrm{2}}=\mathrm{0} \\ $$$${d}\left({x}+{y}\right)^{\mathrm{2}} −\left({x}+{y}\right)×\frac{\mathrm{1}}{\mathrm{2}}{d}\left({y}−\mathrm{1}\right)^{\mathrm{2}} −\frac{{d}\left({y}−\mathrm{1}\right)}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{2}{d}\left({x}+{y}\right)^{\mathrm{2}} −\left({x}+{y}\right){d}\left({y}−\mathrm{1}\right)^{\mathrm{2}} −{d}\left({y}−\mathrm{1}\right)=\mathrm{0} \\ $$$${wait} \\ $$$$ \\ $$

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