By the separation of variables,
For the
integral, we can decompose
![{\displaystyle {\begin{aligned}{\frac {1}{y(y-1)}}={\frac {1}{y-1}}-{\frac {1}{y}},\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d6193c83d8d07f05e5aeb560b372a6475432fc89)
and therefore, we have,
This implies
![{\displaystyle {\begin{aligned}\left|{\frac {y-1}{y}}\right|=e^{C}|x|.\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/80a5454be7caf316e99ae0a5d7477f05ea5ab320)
The initial condition is
, and so we get
. Thus,
![{\displaystyle {\begin{aligned}\left|{\frac {y-1}{y}}\right|=2|x|\end{aligned}}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/af2d3dde7877d730b0018d5e8c7eafa894a92631)
Furthermore, when we use the initial data both quantities are positive and so we can drop the absolute value signs:
![{\displaystyle {\begin{aligned}{\frac {y-1}{y}}&=2x\\1-{\frac {1}{y}}&=2x\\{\frac {1}{y}}&=1-2x\\y&={\frac {1}{1-2x}}.\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/5095d626cc5278b44755dff56f08a8a947087cf3)