The solution of the equation `(dy)/(dx)=(x+y)/(x-y)`, is

Options

A. `C(x^(2)+y^(2))^(1//2)+e^(tan^(-1)((y)/(x)))=0`
B. `C(x^(2)+y^(2))^(1//2)+e^(tan^(-1)((y)/(x)))`
C. `C(x^(2)-y^(2))^(1//2)+e^(tan^(-1)((y)/(x)))`
D. none of these

Answer

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

### Step 1: Identify the Homogeneous Equation
The given equation is homogeneous because both the numerator and denominator are of degree one. We can rewrite the equation in a more manageable form.

### Step 2: Rewrite the Equation
We can express the equation as:
\[ \frac{dy}{dx} = \frac{x+y}{x-y} = \frac{x(1 + \frac{y}{x})}{x(1 - \frac{y}{x})} = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}} \]
Let \(t = \frac{y}{x}\). Then, \(y = tx\).

### Step 3: Differentiate \(y\)
Using the product rule, we differentiate \(y = tx\):
\[ \frac{dy}{dx} = t + x\frac{dt}{dx} \]

### Step 4: Substitute into the Original Equation
Substituting \(y\) and \(\frac{dy}{dx}\) into the original equation gives:
\[ t + x\frac{dt}{dx} = \frac{1 + t}{1 - t} \]

### Step 5: Rearrange the Equation
Rearranging the equation:
\[ x\frac{dt}{dx} = \frac{1 + t}{1 - t} - t \]
Finding a common denominator:
\[ x\frac{dt}{dx} = \frac{1 + t - t(1 - t)}{1 - t} = \frac{1 + t - t + t^2}{1 - t} = \frac{1 + t^2}{1 - t} \]

### Step 6: Separate Variables
We can separate the variables:
\[ \frac{1 - t}{1 + t^2} dt = \frac{1}{x} dx \]

### Step 7: Integrate Both Sides
Now we integrate both sides:
\[ \int \frac{1 - t}{1 + t^2} dt = \int \frac{1}{x} dx \]
The left-hand side can be split into two integrals:
\[ \int \frac{1}{1 + t^2} dt - \int \frac{t}{1 + t^2} dt \]
The first integral is \(\tan^{-1}(t)\) and the second integral can be solved by substitution:
\[ \int \frac{t}{1 + t^2} dt = \frac{1}{2} \ln(1 + t^2) \]
Thus, we have:
\[ \tan^{-1}(t) - \frac{1}{2} \ln(1 + t^2) = \ln|x| + C \]

### Step 8: Substitute Back for \(t\)
Substituting back \(t = \frac{y}{x}\):
\[ \tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2} \ln\left(1 + \left(\frac{y}{x}\right)^2\right) = \ln|x| + C \]

### Step 9: Final Rearrangement
This can be rearranged to express the solution in terms of \(y\) and \(x\):
\[ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C + \frac{1}{2} \ln\left(1 + \frac{y^2}{x^2}\right) \]

### Conclusion
The solution of the differential equation \(\frac{dy}{dx} = \frac{x+y}{x-y}\) can be expressed in the form:
\[ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C + \frac{1}{2} \ln\left(1 + \frac{y^2}{x^2}\right) \]