1. **State the problem:** Solve the differential equation $$xy'' = y' \ln\left(\frac{y'}{x}\right)$$. 2. **Rewrite the equation:** Let $p = y' = \frac{dy}{dx}$ and $p' = y'' = \frac{dp}{dx}$. The equation becomes: $$x p' = p \ln\left(\frac{p}{x}\right)$$ 3. **Rewrite $p'$ in terms of $p$ and $x$:** Since $p = \frac{dy}{dx}$, then $p' = \frac{dp}{dx}$. The equation is: $$x \frac{dp}{dx} = p \ln\left(\frac{p}{x}\right)$$ 4. **Separate variables:** Rewrite as: $$\frac{dp}{dx} = \frac{p}{x} \ln\left(\frac{p}{x}\right)$$ 5. **Substitute $u = \frac{p}{x}$:** Then $p = x u$ and $$\frac{dp}{dx} = u + x \frac{du}{dx}$$ 6. **Substitute into the equation:** $$u + x \frac{du}{dx} = \frac{p}{x} \ln(u) = u \ln(u)$$ 7. **Rearrange:** $$x \frac{du}{dx} = u \ln(u) - u = u (\ln(u) - 1)$$ 8. **Separate variables:** $$\frac{du}{u (\ln(u) - 1)} = \frac{dx}{x}$$ 9. **Integrate both sides:** Let’s integrate the left side with respect to $u$ and the right side with respect to $x$: $$\int \frac{du}{u (\ln(u) - 1)} = \int \frac{dx}{x}$$ 10. **Integrate left side:** Use substitution $w = \ln(u) - 1$, so $dw = \frac{1}{u} du$. Then: $$\int \frac{du}{u (\ln(u) - 1)} = \int \frac{1}{w} dw = \ln|w| + C = \ln|\ln(u) - 1| + C$$ 11. **Integrate right side:** $$\int \frac{dx}{x} = \ln|x| + C$$ 12. **Combine results:** $$\ln|\ln(u) - 1| = \ln|x| + C_1$$ 13. **Exponentiate both sides:** $$|\ln(u) - 1| = C_2 x$$ 14. **Rewrite $u$:** Recall $u = \frac{p}{x} = \frac{y'}{x}$, so: $$|\ln\left(\frac{y'}{x}\right) - 1| = C_2 x$$ 15. **Solve for $y'$:** $$\ln\left(\frac{y'}{x}\right) = 1 \pm C_2 x$$ Exponentiate: $$\frac{y'}{x} = e^{1 \pm C_2 x}$$ So: $$y' = x e^{1 \pm C_2 x}$$ 16. **Integrate to find $y$:** $$y = \int y' dx = \int x e^{1 \pm C_2 x} dx$$ This integral depends on the sign and constant but can be evaluated by integration by parts. **Final answer:** $$y' = x e^{1 + C x}$$ where $C$ is an arbitrary constant. Slug: "differential equation" Subject: "calculus" Desmos: {"latex":"y' = x e^{1 + C x}","features":{"intercepts":true,"extrema":true}} q_count:1