6.9 Example

Solution. We transfer the $(y+1)$ to the left hand side and integrate, to obtain:

and hence ${\log|y+1|=x^{2}-2x+c.}$ To get this into the form $y=h(x)$, we exponentiate both sides to get: $y+1={e^{x^{2}-2x+c},}$ or in other words, $y={Ae^{x^{2}-2x}-1,}$ where $A={e^{c}}$ is a non-zero constant.

Note that we can’t add an arbitrary constant onto $y$ to get another solution to the equation.