We are given the initial value problem (IVP):

$\frac{dy}{dx} = \frac{y (\ln y - \ln x + 1)}{x}, \quad y(1) = e^{-1}$

Step 1: Rearrange the equation

Rearrange the differential equation to isolate terms involving $y$ and $x$:

$\frac{1}{y} \frac{dy}{dx} = \frac{\ln y - \ln x + 1}{x}.$

This suggests that it might be useful to attempt solving by separation of variables.

Step 2: Simplify and integrate both sides

We notice that $\frac{dy}{y}$ is on the left-hand side and the right-hand side involves a logarithmic expression that can potentially be integrated. Let's first simplify and then integrate:

First, separate the equation:

$\frac{1}{y} dy = \frac{(\ln y - \ln x + 1)}{x} dx.$

Now integrate both sides:

$\int \frac{1}{y} dy = \int \frac{\ln y}{x} dx - \int \frac{\ln x}{x} dx + \int \frac{1}{x} dx.$

Step 3: Integrate

• The left-hand side integrates as $\ln y$.
• The right-hand side requires breaking into individual integrals:
  • $\int \frac{\ln y}{x} dx$ will be addressed later (as it’s part of a product).
  • $\int \frac{\ln x}{x} dx = \frac{(\ln x)^2}{2}$.
  • $\int \frac{1}{x} dx = \ln x$.