On putting `(y)/(x)=v` the differential equation `(dy)/(dx)=(2xy-y^(2))/(2xy-x^(2))` is transferred to

Options

A. `x(2v-1)dx=3v(v-1)dx`
B. `x(2v-1)dv=3v(1-v)dx`
C. `x(1-2v)dv=(v^(2)-2v)dx`
D. `(1-2v)dv=(v^(2)-2v)dx`

Explanation

To solve the given differential equation \(\frac{dy}{dx} = \frac{2xy - y^2}{2xy - x^2}\) by substituting \(\frac{y}{x} = v\), we will follow these steps:

### Step 1: Substitute \(y\) in terms of \(v\) and \(x\)
We start with the substitution:
\[ y = vx \]
This implies:
\[ \frac{y}{x} = v \quad \text{or} \quad y = vx \]

### Step 2: Differentiate \(y\) with respect to \(x\)
Using the product rule to differentiate \(y\):
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]

### Step 3: Substitute \(y\) and \(\frac{dy}{dx}\) into the original equation
Now, substituting \(y = vx\) and \(\frac{dy}{dx} = v + x\frac{dv}{dx}\) into the original differential equation:
\[ v + x\frac{dv}{dx} = \frac{2x(vx) - (vx)^2}{2x(vx) - x^2} \]

### Step 4: Simplify the right-hand side
We simplify the right-hand side:
\[ = \frac{2vx^2 - v^2x^2}{2vx^2 - x^2} = \frac{x^2(2v - v^2)}{x^2(2v - 1)} \]
Assuming \(x \neq 0\), we can cancel \(x^2\):
\[ = \frac{2v - v^2}{2v - 1} \]

### Step 5: Set the equation
Now we have:
\[ v + x\frac{dv}{dx} = \frac{2v - v^2}{2v - 1} \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ x\frac{dv}{dx} = \frac{2v - v^2}{2v - 1} - v \]

### Step 7: Combine fractions
To combine the fractions on the right-hand side, we need a common denominator:
\[ x\frac{dv}{dx} = \frac{(2v - v^2) - v(2v - 1)}{2v - 1} \]
This simplifies to:
\[ = \frac{2v - v^2 - 2v^2 + v}{2v - 1} = \frac{3v - 3v^2}{2v - 1} \]

### Step 8: Factor out common terms
Factoring out the common terms gives:
\[ x\frac{dv}{dx} = \frac{3v(1 - v)}{2v - 1} \]

### Step 9: Final form
Thus, we can write:
\[ x(2v - 1) \frac{dv}{dx} = 3v(1 - v) \]

This is the transformed differential equation after substituting \(y/x = v\).

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