The solution of the differential equation, `dy/dx=(x-y)^(2)`,
when `y(1)=1,` is

Options

A. `log_eabs((2-y)/(2-x))=2(y-1)`
B. `log_eabs((2-x)/(2-y))=x-y`
C. `-log_eabs((1+x-y)/(1-x+y))=x+y-2`
D. `-log_eabs((1-x+y)/(1+x-y))=2(x-1)`

Answer

To solve the differential equation \(\frac{dy}{dx} = (x - y)^2\) with the initial condition \(y(1) = 1\), we can follow these steps:

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

### Step 2: Differentiate
Differentiating both sides with respect to \(x\):
\[ \frac{dy}{dx} = 1 - \frac{dt}{dx} \]

### Step 3: Substitute into the Differential Equation
Substituting \(\frac{dy}{dx}\) into the original differential equation:
\[ 1 - \frac{dt}{dx} = t^2 \]
Rearranging gives:
\[ \frac{dt}{dx} = 1 - t^2 \]

### Step 4: Separate Variables
We can separate the variables:
\[ \frac{dt}{1 - t^2} = dx \]

### Step 5: Integrate Both Sides
Integrating both sides:
\[ \int \frac{dt}{1 - t^2} = \int dx \]
The left side can be integrated using the formula for the integral of a rational function:
\[ \frac{1}{2} \ln \left| \frac{1 + t}{1 - t} \right| = x + C \]

### Step 6: Substitute Back for \(t\)
Substituting back \(t = x - y\):
\[ \frac{1}{2} \ln \left| \frac{1 + (x - y)}{1 - (x - y)} \right| = x + C \]

### Step 7: Apply Initial Condition
Using the initial condition \(y(1) = 1\):
\[ \frac{1}{2} \ln \left| \frac{1 + (1 - 1)}{1 - (1 - 1)} \right| = 1 + C \]
This simplifies to:
\[ \frac{1}{2} \ln(1) = 1 + C \implies 0 = 1 + C \implies C = -1 \]

### Step 8: Final Equation
Substituting \(C = -1\) back into the equation:
\[ \frac{1}{2} \ln \left| \frac{1 + (x - y)}{1 - (x - y)} \right| = x - 1 \]
Multiplying both sides by 2:
\[ \ln \left| \frac{1 + (x - y)}{1 - (x - y)} \right| = 2(x - 1) \]

### Step 9: Exponentiate
Exponentiating both sides gives:
\[ \left| \frac{1 + (x - y)}{1 - (x - y)} \right| = e^{2(x - 1)} \]

### Step 10: Solve for \(y\)
This leads to the final solution for \(y\):
\[ \frac{1 + (x - y)}{1 - (x - y)} = e^{2(x - 1)} \]
Cross-multiplying and solving for \(y\) will yield the final form of the solution.

### Summary
The solution to the differential equation \(\frac{dy}{dx} = (x - y)^2\) with the initial condition \(y(1) = 1\) is given by:
\[ \frac{1 + (x - y)}{1 - (x - y)} = e^{2(x - 1)} \]