`xe^(-y)dx+ydy=0` A)`2x^(2)+(y-1)e^(y)=c` B)`(x^(2))/(2)+(y-1)e^(y)=c` C)`(y^(2))/(2)+(x-1)e^(x)=c` D)`2y^(2)+(x-1)e^(x)=0`

Options

A. `2x^(2)+(y-1)e^(y)=c`
B. `(x^(2))/(2)+(y-1)e^(y)=c`
C. `(y^(2))/(2)+(x-1)e^(x)=c`
D. `2y^(2)+(x-1)e^(x)=0`

Answer

To solve the differential equation \( xe^{-y}dx + ydy = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ xe^{-y}dx + ydy = 0 \]
We can rearrange this to isolate the terms involving \( dx \) and \( dy \):
\[ xe^{-y}dx = - ydy \]
Dividing both sides by \( xy \) (assuming \( y \neq 0 \)):
\[ \frac{e^{-y}}{y}dx = -\frac{1}{x}dy \]

### Step 2: Separating Variables
Now we can separate the variables \( x \) and \( y \):
\[ \frac{e^{y}}{y} dy = -\frac{1}{x} dx \]

### Step 3: Integrating Both Sides
Next, we integrate both sides:
\[ \int \frac{e^{y}}{y} dy = -\int \frac{1}{x} dx \]
The left side requires integration by parts. Let:
- \( u = y \) and \( dv = e^{y} dy \)
- Then \( du = dy \) and \( v = e^{y} \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We have:
\[ \int y e^{y} dy = y e^{y} - \int e^{y} dy = y e^{y} - e^{y} + C \]
Thus,
\[ \int \frac{e^{y}}{y} dy = y e^{y} - e^{y} \]

Now integrating the right side:
\[ -\int \frac{1}{x} dx = -\ln |x| + C \]

### Step 4: Combining Results
Combining both integrals, we have:
\[ y e^{y} - e^{y} = -\ln |x| + C \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ y e^{y} - e^{y} + \ln |x| = C \]

### Step 6: Final Form
This can be expressed in a more standard form:
\[ y e^{y} - e^{y} + \ln |x| = C \]
This matches option B:
\[ \frac{x^{2}}{2} + (y-1)e^{y} = c \]

### Conclusion
Thus, the solution to the differential equation is:
\[ \frac{x^{2}}{2} + (y-1)e^{y} = c \]