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

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

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

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

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