The solution of the differential equation `(dy)/(dx)=(x-y)/(x+4y)` is (where C is the constant of integration)

Options

A. `xy+y^(2)=x+C`
B. `xy-y^(2)=x^(2)+C`
C. `xy+2y^(2)=x^(2)+C`
D. `2xy+4y^(2)=x^(2)+C`

Answer

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

### Step 1: Cross Multiply
We start by cross-multiplying to eliminate the fraction:
\[ (x + 4y) dy = (x - y) dx \]

### Step 2: Rearranging the Equation
Next, we rearrange the equation:
\[ x dy + 4y dy = x dx - y dx \]
This can be rewritten as:
\[ x dy + y dx = x dx - 4y dy \]

### Step 3: Recognizing the Product Rule
Notice that the left-hand side can be recognized as the product rule:
\[ d(xy) = x dy + y dx \]
Thus, we can rewrite the equation as:
\[ d(xy) = x dx - 4y dy \]

### Step 4: Integrating Both Sides
Now we integrate both sides:
\[ \int d(xy) = \int (x dx - 4y dy) \]
This gives us:
\[ xy = \int x dx - 4 \int y dy \]

### Step 5: Performing the Integrations
Calculating the integrals:
\[ xy = \frac{x^2}{2} - 4 \cdot \frac{y^2}{2} + C \]
This simplifies to:
\[ xy = \frac{x^2}{2} - 2y^2 + C \]

### Step 6: Rearranging the Equation
To express the solution in a standard form, we can multiply through by 2:
\[ 2xy = x^2 - 4y^2 + 2C \]
Let \(C' = 2C\) (since \(C'\) is also a constant), we have:
\[ 2xy + 4y^2 = x^2 + C' \]

### Final Solution
Thus, the final solution to the differential equation is:
\[ 2xy + 4y^2 = x^2 + C \]