Find the integral of (y2 -1 ) dx-2dy = 0
We have : (y2−1) dx−2 dy=0 ⟹ 2 dyy2−1= dx Integrating both sides , we get ∫2 dyy2−1=∫dx ⟹ ∫(y+1)−(y−1)(y+1)(y−1) dy=∫dx\mathrm{We\;have\;:\;\;}(y^2-1)\;dx-2\;dy=0\\ \;\\ \implies \;\dfrac{2\;dy}{y^2-1}=\;dx\\ \; \\ \textup{Integrating both sides , we get}\\ \; \\ \int \dfrac{2\;dy}{y^2-1}=\int dx\\ \; \\ \implies \;\int \dfrac{(y+1)-(y-1)}{(y+1)(y-1)}\;dy=\int dxWehave:(y2−1)dx−2dy=0⟹y2−12dy=dxIntegrating both sides , we get∫y2−12dy=∫dx⟹∫(y+1)(y−1)(y+1)−(y−1)dy=∫dx
⟹ ∫1y−1 dy−∫1y+1 dy=∫dx ⟹ ln∣y−1∣ −ln∣y+1∣=x+C ⟹ ln∣y−1y+1∣=x+C ⟹ ∣y−1y+1∣=Aex \implies \;\int \dfrac{1}{y-1}\;dy-\int \dfrac{1}{y+1}\;dy=\int dx\\ \;\\ \implies \;\ln |{y-1}|\;-\ln |{y+1}|=x+C\\ \;\\ \implies \;\ln \left|\dfrac{y-1}{y+1}\right|=x+C\\ \;\\ \implies \;\left|\dfrac{y-1}{y+1}\right|=Ae^x\\ \;\\⟹∫y−11dy−∫y+11dy=∫dx⟹ln∣y−1∣−ln∣y+1∣=x+C⟹ln∣∣y+1y−1∣∣=x+C⟹∣∣y+1y−1∣∣=Aex
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