Question

Find the integral of (y² -1 ) dx-2dy = 0

Answer 1

$\mathrm{We\;have\;:\;\;}(y^2-1)\;dx-2\;dy=0\\ \;\\ \implies \;\dfrac{2\;dy}{y^2-1}=\;dx\\ \; \\ \textup{Integrating both sides , we get}\\ \; \\ \int \dfrac{2\;dy}{y^2-1}=\int dx\\ \; \\ \implies \;\int \dfrac{(y+1)-(y-1)}{(y+1)(y-1)}\;dy=\int dx$

$\implies \;\int \dfrac{1}{y-1}\;dy-\int \dfrac{1}{y+1}\;dy=\int dx\\ \;\\ \implies \;\ln |{y-1}|\;-\ln |{y+1}|=x+C\\ \;\\ \implies \;\ln \left|\dfrac{y-1}{y+1}\right|=x+C\\ \;\\ \implies \;\left|\dfrac{y-1}{y+1}\right|=Ae^x\\ \;\\$