The solution of the differential equation
` (dy)/(dx) = (x-2y+1)/(2x -4y)` is

Options

A. `(x-2y)^(2+2x=C`
B. `(x-2y)^(2)+x=C`
C. `(x-2y)+2x^(2)=C`
D. `(x-2y)+x^(2)=C`

Explanation

To solve the differential equation \(\frac{dy}{dx} = \frac{x - 2y + 1}{2x - 4y}\), we will follow these steps:

### Step 1: Substitution
Let \(t = x - 2y\). Then, we differentiate both sides with respect to \(x\):
\[ \frac{dt}{dx} = 1 - 2\frac{dy}{dx} \]

### Step 2: Express \(\frac{dy}{dx}\)
From the above equation, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1 - \frac{dt}{dx}}{2} \]

### Step 3: Substitute into the original equation
Substituting \(t\) into the original differential equation:
\[ \frac{1 - \frac{dt}{dx}}{2} = \frac{t + 1}{2t} \]
Multiplying both sides by 2 gives:
\[ 1 - \frac{dt}{dx} = \frac{t + 1}{t} \]

### Step 4: Rearranging the equation
Rearranging the equation:
\[ 1 - \frac{t + 1}{t} = \frac{dt}{dx} \]
This simplifies to:
\[ \frac{t - 1}{t} = \frac{dt}{dx} \]

### Step 5: Cross-multiplying
Cross-multiplying gives:
\[ t \, dt = (t - 1) \, dx \]

### Step 6: Integrating both sides
Now we integrate both sides:
\[ \int t \, dt = \int (t - 1) \, dx \]
This results in:
\[ \frac{t^2}{2} = tx - x + C \]

### Step 7: Substitute back for \(t\)
Substituting back \(t = x - 2y\):
\[ \frac{(x - 2y)^2}{2} = (x - 2y)x - x + C \]

### Step 8: Rearranging
Rearranging gives:
\[ \frac{(x - 2y)^2}{2} = x^2 - 2xy + C \]

### Step 9: Final form
Multiplying through by 2:
\[ (x - 2y)^2 = 2x^2 - 4xy + 2C \]
Let \(C_1 = 2C\), we can rewrite it as:
\[ (x - 2y)^2 = 2x^2 - 4xy + C_1 \]

### Final Solution
Thus, the solution of the differential equation is:
\[ (x - 2y)^2 - 2x^2 + 4xy = C_1 \]

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