General solution of `e^(x-y)dx+e^(y-x)dy=0` is

Options

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

Answer

To solve the differential equation \( e^{(x-y)}dx + e^{(y-x)}dy = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ e^{(x-y)}dx + e^{(y-x)}dy = 0 \]
This can be rearranged to:
\[ e^{(x-y)}dx = -e^{(y-x)}dy \]

### Step 2: Separate the variables
Now, we can separate the variables by dividing both sides by \( e^{(y-x)} \):
\[ \frac{e^{(x-y)}}{e^{(y-x)}}dx = -dy \]
This simplifies to:
\[ e^{(2x - 2y)}dx = -dy \]

### Step 3: Further simplification
We can rewrite the equation as:
\[ e^{2(x-y)}dx + dy = 0 \]

### Step 4: Integrate both sides
Now, we will integrate both sides. We can rewrite the equation as:
\[ e^{2(x-y)}dx = -dy \]
Integrating both sides, we have:
\[ \int e^{2(x-y)}dx = -\int dy \]

### Step 5: Perform the integration
The left-hand side can be integrated as follows:
\[ \int e^{2(x-y)}dx = \frac{1}{2} e^{2(x-y)} + C_1 \]
The right-hand side integrates to:
\[ -\int dy = -y + C_2 \]

### Step 6: Combine the results
Setting the two integrals equal gives us:
\[ \frac{1}{2} e^{2(x-y)} = -y + C \]
where \( C = C_2 - C_1 \).

### Step 7: Rearranging the equation
Multiplying through by 2, we have:
\[ e^{2(x-y)} = -2y + 2C \]

### Step 8: Expressing the general solution
Let \( C' = 2C \), then we can express the general solution as:
\[ e^{2(x-y)} + 2y = C' \]

### Final General Solution
Thus, the general solution of the differential equation is:
\[ e^{2(x-y)} + 2y = C \]
where \( C \) is an arbitrary constant.

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