To solve the differential equation

dydx=2y−x2x−y,\frac{dy}{dx} = \frac{2y - x}{2x - y},dxdy​=2x−y2y−x​,

To solve the differential equation 

\[
\frac{dy}{dx} = \frac{2y - x}{2x - y},
\]

we can use a substitution method to simplify the equation. Let's proceed step by step.

### Step 1: Check for homogeneity
The given differential equation can be written as:

\[
\frac{dy}{dx} = \frac{2y - x}{2x - y}
\]

We can see that both the numerator and the denominator are linear expressions involving \(x\) and \(y\). This suggests that the equation might be homogeneous, and we can try a substitution of the form:

\[
v = \frac{y}{x} \quad \text{or equivalently} \quad y = vx.
\]

### Step 2: Substitute \( y = vx \) into the equation
Differentiate \( y = vx \) with respect to \( x \):

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

Now, substitute \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \) into the original differential equation:

\[
v + x\frac{dv}{dx} = \frac{2(vx) - x}{2x - vx}.
\]

### Step 3: Simplify the equation
Simplify the right side:

\[
v + x\frac{dv}{dx} = \frac{2vx - x}{2x - vx} = \frac{x(2v - 1)}{x(2 - v)}.
\]

The equation simplifies to:

\[
v + x\frac{dv}{dx} = \frac{2v - 1}{2 - v}.
\]

### Step 4: Separate variables and integrate
Rearrange the equation to separate variables:

\[
x\frac{dv}{dx} = \frac{2v - 1}{2 - v} - v.
\]

Simplify the right-hand side:

\[
x\frac{dv}{dx} = \frac{2v - 1 - v(2 - v)}{2 - v} = \frac{2v - 1 - 2v + v^2}{2 - v} = \frac{v^2 - 1}{2 - v}.
\]

This gives:

\[
\frac{2 - v}{v^2 - 1} dv = \frac{dx}{x}.
\]

### Step 5: Integrate both sides
Now integrate both sides:

\[
\int \frac{2 - v}{v^2 - 1} dv = \int \frac{dx}{x}.
\]

You can break the integral on the left side into two parts and integrate:

\[
\int \left(\frac{2}{v^2 - 1} - \frac{v}{v^2 - 1}\right) dv = \int \frac{dx}{x}.
\]

The first integral can be split as follows:

\[
\int \frac{2}{(v-1)(v+1)} dv =  \int \left( \frac{1}{v-1} - \frac{1}{v+1} \right) dv,
\]

and the second integral is:

\[
\int \frac{v}{v^2 - 1} dv = \frac{1}{2}\ln|v^2 - 1|.
\]

So the integrals become:

\[
\ln\left|\frac{v-1}{v+1}\right| - \frac{1}{2}\ln|v^2 - 1| = \ln|x| + C.
\]

Combine the logarithms and back-substitute \( v = \frac{y}{x} \) to find the general solution.

### Step 6: Back-substitute and solve for \( y \)
After back-substitution, simplify the expression to get \( y \) in terms of \( x \) and \( C \).

---

Would you like to go through the details of each integration step? 

Here are 5 relative questions that might interest you:

1. How can you determine if a differential equation is homogeneous?
2. What other methods can solve first-order differential equations besides substitution?
3. How does the method of integrating factors work in solving differential equations?
4. What are exact differential equations, and how can they be identified?
5. Can you always use the substitution \( v = \frac{y}{x} \) for any homogeneous equation?

**Tip:** When dealing with homogeneous differential equations, always check if a substitution like \( y = vx \) simplifies the problem significantly.

To solve the differential equation \(\frac{dy}{dx} = \frac{2y - x}{2x - y}\),

Explore the step-by-step solution to solving the homogeneous differential equation \(\frac{dy}{dx} = \frac{2y - x}{2x - y}\). This guide covers substitution methods, separation of variables, and integration techniques for rational functions. Ideal for undergraduate mathematics students.

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