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Solve dy=(y²-1)dx.
Ans: ln((y-1)/(y+1))=2x+c

User MKaras
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1 Answer

4 votes

Divide both sides by
y^2-1 and integrate:


(\mathrm dy)/(y^2-1)=\mathrm dx

To integrate left side, first expand into partial fraction:


\frac1{y^2-1}=\frac12\left(\frac1{y-1}-\frac1{y+1}\right)


\displaystyle\int(\mathrm dy)/(y^2-1)=\int\mathrm dx


\frac12\left(\ln|y-1|-\ln|y+1|\right)=x+C


\ln\left|(y-1)/(y+1)\right|=2x+C

User Kerwin
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