The equation of the curve passing through the origin and satisfying the differential equation `((dy)/(dx))^(2)=(x-y)^(2)`, is

Options

A. `e^(2x) (1-x+y)=1+x-y`
B. `e^(2x) (1 +x -y)=1-x+y`
C. `e^(2x)(1-x+y)=1+x+y`
D. `e^2x (1+x+y)=1-x+y`

Answer

To solve the differential equation \(\left(\frac{dy}{dx}\right)^2 = (x - y)^2\), we will follow a systematic approach.

### Step 1: Rewrite the equation
We start with the given equation:
\[ \left(\frac{dy}{dx}\right)^2 = (x - y)^2 \]
Taking the square root on both sides, we have:
\[ \frac{dy}{dx} = \pm (x - y) \]

### Step 2: Solve the positive case
Let's first consider the positive case:
\[ \frac{dy}{dx} = x - y \]
Rearranging gives:
\[ dy + y = x \, dx \]
This is a linear first-order differential equation.

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int 1 \, dx} = e^x \]

### Step 4: Multiply the equation by the integrating factor
Multiplying the entire equation by \( e^x \):
\[ e^x dy + e^x y = e^x x \, dx \]

### Step 5: Recognize the left side as a derivative
The left side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(e^x y) = e^x x \]

### Step 6: Integrate both sides
Integrating both sides with respect to \( x \):
\[ e^x y = \int e^x x \, dx \]
Using integration by parts, let \( u = x \) and \( dv = e^x dx \):
\[ du = dx, \quad v = e^x \]
Thus,
\[ \int e^x x \, dx = e^x x - \int e^x \, dx = e^x x - e^x + C \]

### Step 7: Solve for \( y \)
Substituting back, we have:
\[ e^x y = e^x x - e^x + C \]
Dividing by \( e^x \):
\[ y = x - 1 + Ce^{-x} \]

### Step 8: Apply the initial condition
Since the curve passes through the origin (0,0), we substitute \( x = 0 \) and \( y = 0 \):
\[ 0 = 0 - 1 + C \implies C = 1 \]
Thus, the equation becomes:
\[ y = x - 1 + e^{-x} \]

### Step 9: Solve the negative case
Now consider the negative case:
\[ \frac{dy}{dx} = -(x - y) \]
Rearranging gives:
\[ dy + y = -x \, dx \]
Following similar steps as above, we find:
\[ e^x y = -\int e^x x \, dx \]
This leads to:
\[ y = -x + 1 - Ce^{-x} \]
Applying the initial condition again gives \( C = 1 \), leading to:
\[ y = -x + 1 - e^{-x} \]

### Final Solution
Thus, the general solution of the differential equation is:
\[ y = x - 1 + e^{-x} \quad \text{or} \quad y = -x + 1 - e^{-x} \]