Solve the differential equation `ydx+ {xlog ((y)/(x))} dy-2x dy=0`

Options

A. `y=log (xy+1) +c, x ne 0`
B. `cy= log (x^(2) + y^(2)), x ne 0`
C. `y= log (xy-1)+c, x ne 0`
D. `cy = log ((y)/(x))+1, x ne 0`

Answer

To solve the differential equation \( y \, dx + \left( x \log \left( \frac{y}{x} \right) - 2x \right) dy = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ y \, dx + \left( x \log \left( \frac{y}{x} \right) - 2x \right) dy = 0 \]
This can be rearranged to:
\[ y \, dx = -\left( x \log \left( \frac{y}{x} \right) - 2x \right) dy \]

### Step 2: Divide by \( dy \)
Dividing the entire equation by \( dy \) gives:
\[ \frac{dx}{dy} + \frac{x \log \left( \frac{y}{x} \right) - 2x}{y} = 0 \]
This simplifies to:
\[ \frac{dx}{dy} + \frac{x}{y} \log \left( \frac{y}{x} \right) = \frac{2x}{y} \]

### Step 3: Substitute \( t = \frac{x}{y} \)
Let \( t = \frac{x}{y} \), then \( x = ty \) and \( dx = y \, dt + t \, dy \). Substituting these into the equation gives:
\[ y \left( y \frac{dt}{dy} + t \right) + \frac{ty}{y} \log \left( \frac{y}{ty} \right) = 2t \]
This simplifies to:
\[ y^2 \frac{dt}{dy} + ty + t \log \left( \frac{1}{t} \right) = 2t \]

### Step 4: Simplify the equation
Rearranging gives:
\[ y^2 \frac{dt}{dy} = 2t - ty - t \log t \]
Factoring out \( t \):
\[ y^2 \frac{dt}{dy} = t(2 - y - \log t) \]

### Step 5: Separate variables
Separating variables yields:
\[ \frac{dt}{t(2 - y - \log t)} = \frac{dy}{y^2} \]

### Step 6: Integrate both sides
Integrate both sides:
\[ \int \frac{dt}{t(2 - y - \log t)} = \int \frac{dy}{y^2} \]

### Step 7: Solve the integrals
The left side requires partial fraction decomposition, while the right side integrates to:
\[ -\frac{1}{y} + C \]

### Step 8: Back-substitute \( t \)
After integrating, we will back-substitute \( t = \frac{x}{y} \) into our solution.

### Final Solution
The final solution can be expressed in terms of \( x \) and \( y \) after simplifying the integrals and back-substituting.

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