1. **State the problem:** Given the differential equation $y y'' = y'' - y'$, we want to simplify and solve this or analyze it. 2. **Rewrite the equation:** $$y y'' = y'' - y'$$ 3. **Bring all terms to one side:** $$y y'' - y'' + y' = 0$$ 4. **Factor terms involving $y''$:** $$y''(y - 1) + y' = 0$$ 5. **Isolate $y''$:** $$y''(y - 1) = - y'$$ 6. **If $y \neq 1$, then:** $$y'' = \frac{- y'}{y - 1}$$ 7. **This is a second-order nonlinear differential equation for $y(x)$.** To solve it, we can set $p = y'$, then $p' = y''$. Rewrite as: $$p' = -\frac{p}{y - 1}$$ 8. **Using chain rule for $p' = \frac{dp}{dx} = \frac{dp}{dy} \cdot \frac{dy}{dx} = p \frac{dp}{dy}$, we have:** $$p \frac{dp}{dy} = -\frac{p}{y - 1}$$ 9. **Divide both sides by $p$ (assuming $p \neq 0$):** $$\frac{dp}{dy} = -\frac{1}{y - 1}$$ 10. **Integrate both sides with respect to $y$:** $$\int dp = -\int \frac{1}{y - 1} dy$$ 11. **This gives:** $$p = - \ln|y - 1| + C$$ 12. **Recall $p = \frac{dy}{dx}$, so:** $$\frac{dy}{dx} = C - \ln|y - 1|$$ 13. **This is an implicit first-order differential equation relating $x$ and $y$. Further exact solutions depend on initial conditions or numerical methods.** **Final answer:** The equation reduces to $$y'' = -\frac{y'}{y-1}$$ and further to $$\frac{dy}{dx} = C - \ln|y - 1|.$$