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

Options

A. `- log |( 1 + X-y)/( 1- x+y)|= x + y=2`
B. ` log _(e) |( 2-y) /( 2-x) | = 2 (y-1)`
C. `- log _(e) |(1-x +y) /( 1+ x -y) |=2 (x-1)`
D. ` log _(e) |(2-x )/( 2-y) |= x-y`

Answer

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

### Step 1: Substitute \(u = x - y\)
Let \(u = x - y\). Then, we can express \(y\) in terms of \(u\):
\[ y = x - u \]
Now, differentiate both sides with respect to \(x\):
\[ \frac{dy}{dx} = 1 - \frac{du}{dx} \]

### Step 2: Rewrite the differential equation
Substituting \(\frac{dy}{dx}\) into the original equation gives:
\[ 1 - \frac{du}{dx} = u^2 \]
Rearranging this, we have:
\[ \frac{du}{dx} = 1 - u^2 \]

### Step 3: Separate variables
We can separate the variables:
\[ \frac{du}{1 - u^2} = dx \]

### Step 4: Integrate both sides
Now we integrate both sides. The left side requires partial fraction decomposition:
\[ \frac{1}{1 - u^2} = \frac{1}{2(1-u)} + \frac{1}{2(1+u)} \]
Thus, we have:
\[ \int \frac{du}{1 - u^2} = \frac{1}{2} \int \left( \frac{1}{1-u} + \frac{1}{1+u} \right) du = \frac{1}{2} \left( -\log|1-u| + \log|1+u| \right) + C \]
This simplifies to:
\[ \frac{1}{2} \log \left| \frac{1+u}{1-u} \right| = x + C \]

### Step 5: Solve for \(u\)
Exponentiating both sides gives:
\[ \frac{1+u}{1-u} = e^{2(x+C)} = e^{2x} e^{2C} \]
Let \(k = e^{2C}\), then:
\[ \frac{1+u}{1-u} = ke^{2x} \]
Cross-multiplying yields:
\[ 1 + u = ke^{2x}(1 - u) \]
This simplifies to:
\[ 1 + u = ke^{2x} - ke^{2x}u \]
Rearranging gives:
\[ u(1 + ke^{2x}) = ke^{2x} - 1 \]
Thus:
\[ u = \frac{ke^{2x} - 1}{1 + ke^{2x}} \]

### Step 6: Substitute back for \(y\)
Recall that \(u = x - y\), so:
\[ x - y = \frac{ke^{2x} - 1}{1 + ke^{2x}} \]
Therefore:
\[ y = x - \frac{ke^{2x} - 1}{1 + ke^{2x}} \]

### Step 7: Apply the initial condition \(y(1) = 1\)
Substituting \(x = 1\) and \(y = 1\):
\[ 1 = 1 - \frac{ke^{2 \cdot 1} - 1}{1 + ke^{2 \cdot 1}} \]
This simplifies to:
\[ 0 = -\frac{ke^{2} - 1}{1 + ke^{2}} \]
Thus, \(ke^{2} - 1 = 0\) implies \(k = \frac{1}{e^{2}}\).

### Final Solution
Substituting \(k\) back into the equation for \(y\):
\[ y = x - \frac{\frac{1}{e^{2}} e^{2x} - 1}{1 + \frac{1}{e^{2}} e^{2x}} = x - \frac{e^{2x-2} - 1}{1 + e^{2x-2}} = x - \frac{e^{2x-2} - 1}{e^{2x-2} + 1} \]

This is the solution to the differential equation with the given initial condition.