1. **State the problem:** Find the general solution of the differential equation $$y' = \ln x$$. 2. **Recall the formula:** Since $$y' = \frac{dy}{dx}$$, the equation means $$\frac{dy}{dx} = \ln x$$. 3. **Integrate both sides:** To find $$y$$, integrate $$\ln x$$ with respect to $$x$$: $$y = \int \ln x \, dx + C$$ where $$C$$ is the constant of integration. 4. **Use integration by parts:** Recall the formula: $$\int u \, dv = uv - \int v \, du$$ Let: - $$u = \ln x \implies du = \frac{1}{x} dx$$ - $$dv = dx \implies v = x$$ 5. **Apply integration by parts:** $$\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - \int 1 \, dx$$ 6. **Simplify the integral:** $$\int 1 \, dx = x$$ So, $$\int \ln x \, dx = x \ln x - x + C$$ 7. **Write the general solution:** $$y = x \ln x - x + C$$ This is the general solution to the differential equation $$y' = \ln x$$.