Find the the solution of the differential equations... xy' + y ln(x) = y ln(y) + y

The given differential equation is:

\[
x \frac{dy}{dx} + y \ln(x) = y \ln(y) + y
\]

### Step 1: Simplify the equation
First, divide the entire equation by \( y \) (assuming \( y \neq 0 \)):

\[
\frac{x}{y} \frac{dy}{dx} + \ln(x) = \ln(y) + 1
\]

### Step 2: Separate the variables
Rearrange the terms to separate the variables:

\[
\frac{x}{y} \frac{dy}{dx} = \ln(y) - \ln(x) + 1
\]

Recognize that \( \ln(y) - \ln(x) = \ln\left(\frac{y}{x}\right) \), so the equation becomes:

\[
\frac{x}{y} \frac{dy}{dx} = \ln\left(\frac{y}{x}\right) + 1
\]

### Step 3: Change of variables
Let's introduce a substitution to simplify the equation. Define:

\[
v = \frac{y}{x}
\]

Then, \( y = vx \), and differentiating with respect to \( x \):

\[
\frac{dy}{dx} = v + x \frac{dv}{dx}
\]

Substitute into the equation:

\[
\frac{x}{vx} \left(v + x \frac{dv}{dx}\right) = \ln(v) + 1
\]

Simplify:

\[
\frac{1}{v}(v + x \frac{dv}{dx}) = \ln(v) + 1
\]

\[
1 + \frac{x}{v} \frac{dv}{dx} = \ln(v) + 1
\]

Subtract 1 from both sides:

\[
\frac{x}{v} \frac{dv}{dx} = \ln(v)
\]

### Step 4: Separate and integrate
Separate the variables:

\[
\frac{1}{\ln(v)} dv = \frac{1}{x} dx
\]

Now, integrate both sides:

\[
\int \frac{1}{\ln(v)} dv = \int \frac{1}{x} dx
\]

The integral on the right-hand side is straightforward:

\[
\int \frac{1}{x} dx = \ln|x| + C_1
\]

The integral on the left-hand side is a special function, known as the logarithmic integral, but in a simpler form it can be solved as:

\[
\text{Li}(v) = \ln|x| + C
\]

### Final Answer:
The general solution of the differential equation is:
In implicit form using special functions.

Would you like more details on this, or other questions:

1. How to calculate special functions?
2. Alternative methods for solution
3. Particular conditions
4. Functions relations with integral result.


Find the solution of the differential equation: xy' + y ln(x) = y ln(y) + y

Explore the step-by-step solution to the differential equation xy' + y ln(x) = y ln(y) + y. This problem involves key concepts in differential equations, including change of variables, logarithmic functions, and separation of variables. Suitable for undergraduate students.

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