Solution of the equation `(dy)/(dx) = e^(x-y) (e^(x)-e^(y))` is equal to

Options

A. `e^(y) =e^(x)-1+ce^(-e^(x))`
B. `e^(y-x)= -1 +ce^(-x)`
C. `e^(x)+e^(y)=ce^(e^(x))`
D. none of these

Answer

To solve the differential equation

\[ \frac{dy}{dx} = e^{x-y} (e^x - e^y), \]

we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation in a more manageable form:

\[ \frac{dy}{dx} = e^{x-y} (e^x - e^y). \]

### Step 2: Expand the right-hand side
We can expand the right-hand side:

\[ \frac{dy}{dx} = e^x e^{x-y} - e^y e^{x-y}. \]

This simplifies to:

\[ \frac{dy}{dx} = e^{2x} e^{-y} - e^{x} e^{x-y}. \]

### Step 3: Rearranging the equation
Rearranging gives us:

\[ \frac{dy}{dx} + e^x e^{x-y} = e^{2x} e^{-y}. \]

### Step 4: Substitute \( v = e^y \)
Let’s make a substitution \( v = e^y \), which implies \( y = \ln v \) and consequently, \( \frac{dy}{dx} = \frac{1}{v} \frac{dv}{dx} \).

Substituting this into our equation gives:

\[ \frac{1}{v} \frac{dv}{dx} + e^x \frac{v}{e^y} = e^{2x} \frac{1}{v}. \]

### Step 5: Simplifying the equation
Substituting \( e^y = v \):

\[ \frac{1}{v} \frac{dv}{dx} + e^x \frac{v}{v} = e^{2x} \frac{1}{v}. \]

This simplifies to:

\[ \frac{dv}{dx} + e^x v = e^{2x}. \]

### Step 6: Identify the linear form
This is now a linear first-order differential equation of the form:

\[ \frac{dv}{dx} + p(x)v = q(x), \]

where \( p(x) = e^x \) and \( q(x) = e^{2x} \).

### Step 7: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int p(x)dx} = e^{\int e^x dx} = e^{e^x}. \]

### Step 8: Multiply through by the integrating factor
Multiplying the entire equation by the integrating factor:

\[ e^{e^x} \frac{dv}{dx} + e^{e^x} e^x v = e^{e^x} e^{2x}. \]

### Step 9: Left-hand side as a derivative
The left-hand side can be expressed as the derivative of a product:

\[ \frac{d}{dx}(e^{e^x} v) = e^{e^x + 2x}. \]

### Step 10: Integrate both sides
Integrating both sides gives:

\[ e^{e^x} v = \int e^{e^x + 2x} dx + C. \]

### Step 11: Solve for \( v \)
Now, we can solve for \( v \):

\[ v = e^{-e^x} \left( \int e^{e^x + 2x} dx + C \right). \]

### Step 12: Substitute back for \( y \)
Since \( v = e^y \), we have:

\[ e^y = e^{-e^x} \left( \int e^{e^x + 2x} dx + C \right). \]

Taking the natural logarithm gives us:

\[ y = -e^x + \ln\left( \int e^{e^x + 2x} dx + C \right). \]

### Final Solution
Thus, the solution of the differential equation is:

\[ y = -e^x + \ln\left( \int e^{e^x + 2x} dx + C \right). \]