1. **State the problem:** Solve the differential equation $$\frac{dy}{dx} \ln x + y = e^x$$ for $y$ as a function of $x$.\n\n2. **Rewrite the equation:** The equation is $$\frac{dy}{dx} \ln x + y = e^x.$$ We want to isolate $\frac{dy}{dx}$ or rewrite it in a standard form.\n\n3. **Divide both sides by $\ln x$ (assuming $x>0$ and $x \neq 1$ so $\ln x \neq 0$):**\n$$\frac{dy}{dx} + \frac{y}{\ln x} = \frac{e^x}{\ln x}.$$\nThis is a first-order linear differential equation of the form $$\frac{dy}{dx} + P(x)y = Q(x)$$ where $$P(x) = \frac{1}{\ln x}$$ and $$Q(x) = \frac{e^x}{\ln x}.$$\n\n4. **Find the integrating factor (IF):**\n$$IF = e^{\int P(x) dx} = e^{\int \frac{1}{\ln x} dx}.$$\nThis integral does not have a simple elementary form, so we keep it as is for now.\n\n5. **Express the solution formally:**\nThe solution is given by\n$$y \cdot IF = \int Q(x) \cdot IF \, dx + C,$$\nwhere $C$ is the constant of integration.\n\n6. **Summary:** The solution to the differential equation is\n$$y = \frac{1}{IF} \left( \int \frac{e^x}{\ln x} IF \, dx + C \right),$$\nwhere $$IF = e^{\int \frac{1}{\ln x} dx}.$$\n\nSince the integral $$\int \frac{1}{\ln x} dx$$ does not have a closed form in elementary functions, the solution is expressed in terms of this integral and the constant $C$.