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

Options

A. ` Y=(1)/(2) log |(x-y-1)/(x-y-1) |+C`
B. ` Y=(1)/(2) log |(x+y-1)/(x+y+1) |+C`
C. both a and B
D. none of these

Answer

To solve the differential equation \((x - y)^2 \frac{dy}{dx} = a^2\), we will follow these steps:

### Step 1: Substitute \(t = x - y\)
Let \(t = x - y\). Then, we can express \(y\) in terms of \(t\) and \(x\):
\[ y = x - t \]

### Step 2: Differentiate \(t\) with respect to \(x\)
Differentiating both sides with respect to \(x\):
\[ \frac{dt}{dx} = 1 - \frac{dy}{dx} \]
Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = 1 - \frac{dt}{dx} \]

### Step 3: Substitute into the original equation
Substituting \(t\) and \(\frac{dy}{dx}\) back into the original equation:
\[ t^2 \left(1 - \frac{dt}{dx}\right) = a^2 \]
Expanding this gives:
\[ t^2 - t^2 \frac{dt}{dx} = a^2 \]
Rearranging leads to:
\[ t^2 \frac{dt}{dx} = t^2 - a^2 \]

### Step 4: Separate variables
We can separate the variables:
\[ \frac{dt}{t^2 - a^2} = \frac{dx}{t^2} \]

### Step 5: Integrate both sides
Now we will integrate both sides:
\[ \int \frac{dt}{t^2 - a^2} = \int \frac{dx}{t^2} \]
The left side can be integrated using partial fractions:
\[ \int \left(\frac{1}{2a} \left(\frac{1}{t - a} - \frac{1}{t + a}\right)\right) dt = \int \frac{dx}{t^2} \]
The right side integrates to:
\[ -\frac{1}{t} + C \]

### Step 6: Combine results
Combining the results from both integrations gives:
\[ \frac{1}{2a} \ln \left|\frac{t - a}{t + a}\right| = -\frac{1}{t} + C \]

### Step 7: Substitute back for \(t\)
Now substitute back \(t = x - y\):
\[ \frac{1}{2a} \ln \left|\frac{x - y - a}{x - y + a}\right| = -\frac{1}{x - y} + C \]

### Final Solution
Thus, the solution of the differential equation is:
\[ \frac{1}{2a} \ln \left|\frac{x - y - a}{x - y + a}\right| = -\frac{1}{x - y} + C \]

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