If `y'=(x-y)/(x+y)` , then its solution is

Options

A. `y^2+2xy-x^2=c `
B. `y^2+2xy+x^2=c `
C. `y^2-2xy-x^2=c`
D. `y^2-2xy+x^2=c `

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{dy}{dx} = \frac{x - y}{x + y} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ (x + y) dy = (x - y) dx \]

### Step 3: Rearranging the terms
Rearranging the terms, we get:
\[ x dy + y dy + y dx = x dx \]
This can be rewritten as:
\[ y dy + x dy + y dx = x dx \]

### Step 4: Grouping the terms
Now, we can group the terms:
\[ y dy = x dx - x dy \]

### Step 5: Isolate dy
Rearranging gives us:
\[ y dy + x dy = x dx \]
Factoring out \( dy \):
\[ dy(y + x) = x dx \]

### Step 6: Separate variables
Now we can separate the variables:
\[ \frac{dy}{y + x} = \frac{x}{x} dx \]

### Step 7: Integrate both sides
Integrating both sides:
\[ \int \frac{dy}{y + x} = \int dx \]
This gives us:
\[ \ln |y + x| = x + C \]

### Step 8: Exponentiate to eliminate the logarithm
Exponentiating both sides results in:
\[ |y + x| = e^{x + C} = e^C e^x \]
Let \( k = e^C \), so we have:
\[ y + x = k e^x \]

### Step 9: Solve for y
Finally, we can solve for \( y \):
\[ y = k e^x - x \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = k e^x - x \]