Solution of the differential equation `(dy)/(dx)+(x-2y)/(2x-y)=0` is

Options

A. `(x+y)^(3)=c(y-x)`
B. `(x+y)^(3)=c(x-y)`
C. `(x-y)^(3)=c(x+y)`
D. `(x-y)^(3)=c(x+y)^(2)`

Answer

To solve the differential equation

\[ \frac{dy}{dx} + \frac{x - 2y}{2x - y} = 0, \]

we can follow these steps:

### Step 1: Rearranging the Equation
First, we can rearrange the equation to isolate \(\frac{dy}{dx}\):

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

### Step 2: Substitution
Next, we will use the substitution \(y = vx\), where \(v\) is a function of \(x\). Therefore, we have:

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

Substituting \(y = vx\) into the equation gives:

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

### Step 3: Simplifying the Right Side
Now, simplify the right-hand side:

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

Thus, we can rewrite the equation as:

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

### Step 4: Rearranging the Equation
Rearranging gives:

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

### Step 5: Combining Terms
Combine the terms on the right-hand side:

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

### Step 6: Separating Variables
Now, we can separate the variables:

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

### Step 7: Integrating Both Sides
Integrate both sides:

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

### Step 8: Solving the Integrals
The left side can be split into two integrals:

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

Using the integral formulas, we find:

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

And the right side gives:

\[ -\ln |x| + C. \]

### Step 9: Combining Results
Combining the results, we have:

\[ \ln |1 - v^2| = -\ln |x| + C. \]

### Step 10: Exponentiating
Exponentiating both sides gives:

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

### Step 11: Back Substituting for \(y\)
Recall that \(v = \frac{y}{x}\), so we can substitute back:

\[ |1 - \left(\frac{y}{x}\right)^2| = \frac{C}{|x|}. \]

### Step 12: Final Form
This leads to the final form of the solution:

\[ |y^2 - x^2| = Cx. \]

### Conclusion
Thus, the solution of the given differential equation is:

\[ y^2 - x^2 = Cx, \]

where \(C\) is a constant.

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