Question

dy/dx-y=y^2-1

Answer 1

\[
\frac{dy}{dx} = y^2+y-1
\]So \[
\frac{1}{y^2+y-1}\frac{dy}{dx} = 1
\]Integrate by \(x\): \[
\int \frac{1}{y^2+y-1}\frac{dy}{dx}dx = \int 1dx
\]Remember that \(dy = (dy/dx) dx\).
This means \[
\int \frac{1}{y^2+y-1}dy=\int dx
\]