Solution of the differential equation
`x(dy)/(dx) +y log y=xy e^(x)` is
`x log y = e^(x) (x-1) +c`,

Answer

To solve the differential equation \( x \frac{dy}{dx} + y \log y = xy e^x \), we will follow these steps:

### Step 1: Rewrite the equation
First, we can rewrite the given differential equation in a more manageable form. We divide the entire equation by \( xy \):

\[ \frac{1}{y} \frac{dy}{dx} + \frac{\log y}{x} = e^x \]

### Step 2: Substitution
Next, we can use a substitution to simplify the equation. Let \( v = \log y \). Then, we have:

\[ y = e^v \quad \text{and} \quad \frac{dy}{dx} = e^v \frac{dv}{dx} \]

Substituting these into the equation gives:

\[ \frac{1}{e^v} e^v \frac{dv}{dx} + \frac{v}{x} = e^x \]

This simplifies to:

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

### Step 3: Identify the linear differential equation
The equation \( \frac{dv}{dx} + \frac{v}{x} = e^x \) is a linear differential equation in the standard form:

\[ \frac{dv}{dx} + P(x)v = Q(x) \]

where \( P(x) = \frac{1}{x} \) and \( Q(x) = e^x \).

### Step 4: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\log x} = x \]

### Step 5: Multiply through by the integrating factor
We multiply the entire differential equation by the integrating factor \( x \):

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

### Step 6: Solve the equation
The left-hand side can be rewritten as the derivative of a product:

\[ \frac{d}{dx}(xv) = x e^x \]

Integrating both sides gives:

\[ xv = \int x e^x \, dx \]

To solve the integral on the right, we use integration by parts:

Let \( u = x \) and \( dv = e^x \, dx \). Then \( du = dx \) and \( v = e^x \).

Using integration by parts:

\[ \int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C \]

Thus, we have:

\[ xv = x e^x - e^x + C \]

### Step 7: Solve for \( v \)
Now, we solve for \( v \):

\[ v = e^x - \frac{e^x}{x} + \frac{C}{x} \]

Recalling that \( v = \log y \), we substitute back:

\[ \log y = e^x - \frac{e^x}{x} + \frac{C}{x} \]

### Step 8: Exponentiate to find \( y \)
Exponentiating both sides gives:

\[ y = e^{e^x - \frac{e^x}{x} + \frac{C}{x}} = e^{e^x} \cdot e^{-\frac{e^x}{x}} \cdot e^{\frac{C}{x}} \]

### Final Step: Rearranging the solution
We can rearrange the solution to match the form given in the question:

\[ x \log y = e^x (x - 1) + C \]

### Conclusion
Thus, we have shown that the solution of the differential equation is indeed:

\[ x \log y = e^x (x - 1) + C \]