1. **State the problem:** We need to find the general solution of the differential equation $$ yy'' = (y')^2 (1 - y' \sin y - y y' \cos y) $$ 2. **Rewrite the equation:** Let $p = y' = \frac{dy}{dx}$ and $p' = y'' = \frac{dp}{dx}$. The equation becomes $$ y p' = p^2 (1 - p \sin y - y p \cos y) $$ 3. **Express $p'$ in terms of $y$ and $p$:** Using the chain rule, $$ p' = \frac{dp}{dx} = \frac{dp}{dy} \frac{dy}{dx} = p \frac{dp}{dy} $$ Substitute into the equation: $$ y p \frac{dp}{dy} = p^2 (1 - p \sin y - y p \cos y) $$ 4. **Simplify:** Divide both sides by $p$ (assuming $p \neq 0$): $$ y \frac{dp}{dy} = p (1 - p \sin y - y p \cos y) $$ 5. **Rewrite:** $$ y \frac{dp}{dy} = p - p^2 \sin y - y p^2 \cos y $$ 6. **Bring all terms to one side:** $$ y \frac{dp}{dy} - p + p^2 \sin y + y p^2 \cos y = 0 $$ 7. **Divide by $y$ (assuming $y \neq 0$):** $$ \frac{dp}{dy} - \frac{p}{y} + \frac{p^2}{y} \sin y + p^2 \cos y = 0 $$ 8. **Rewrite as:** $$ \frac{dp}{dy} = \frac{p}{y} - \frac{p^2}{y} \sin y - p^2 \cos y $$ 9. **This is a Bernoulli differential equation in $p(y)$:** $$ \frac{dp}{dy} + \left( \frac{\sin y}{y} + \cos y \right) p^2 = \frac{p}{y} $$ 10. **Divide entire equation by $p^2$ to transform:** Let $z = \frac{1}{p}$, then $$ \frac{dp}{dy} = -\frac{1}{z^2} \frac{dz}{dy} $$ Substitute into the equation: $$ -\frac{1}{z^2} \frac{dz}{dy} + \left( \frac{\sin y}{y} + \cos y \right) \frac{1}{z^2} = \frac{1}{y z} $$ Multiply both sides by $z^2$: $$ -\frac{dz}{dy} + \left( \frac{\sin y}{y} + \cos y \right) = \frac{z}{y} $$ 11. **Rearranged:** $$ \frac{dz}{dy} + \left(-\frac{1}{y}\right) z = \frac{\sin y}{y} + \cos y $$ 12. **This is a linear first-order ODE in $z(y)$:** $$ \frac{dz}{dy} + P(y) z = Q(y) $$ where $$ P(y) = -\frac{1}{y}, \quad Q(y) = \frac{\sin y}{y} + \cos y $$ 13. **Find integrating factor (IF):** $$ IF = e^{\int P(y) dy} = e^{-\int \frac{1}{y} dy} = e^{-\ln |y|} = \frac{1}{y} $$ 14. **Multiply both sides by IF:** $$ \frac{1}{y} \frac{dz}{dy} - \frac{1}{y^2} z = \frac{\sin y}{y^2} + \frac{\cos y}{y} $$ This is equivalent to $$ \frac{d}{dy} \left( \frac{z}{y} \right) = \frac{\sin y}{y^2} + \frac{\cos y}{y} $$ 15. **Integrate both sides:** $$ \frac{z}{y} = \int \left( \frac{\sin y}{y^2} + \frac{\cos y}{y} \right) dy + C $$ 16. **Recall $z = \frac{1}{p} = \frac{1}{y'}$, so** $$ \frac{1}{y'} = y \left( \int \left( \frac{\sin y}{y^2} + \frac{\cos y}{y} \right) dy + C \right) $$ 17. **This implicit form represents the general solution.** **Summary:** The general solution is given implicitly by $$ \frac{1}{y'} = y \left( \int \left( \frac{\sin y}{y^2} + \frac{\cos y}{y} \right) dy + C \right) $$ where $C$ is an arbitrary constant. This completes the solution.