The solution of the differential equation `xdy + ydx= xydx` when y(1)=1 is

Options

A. `y= (e^(x))/(x)`
B. `y= (e^(x))/(ex)`
C. `y= (xe^(x))/(e )`
D. None of these

Answer

To solve the differential equation \( x \, dy + y \, dx = xy \, dx \) with the initial condition \( y(1) = 1 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ x \, dy + y \, dx = xy \, dx \]
We can rearrange this to isolate \( dy \):
\[ x \, dy = xy \, dx - y \, dx \]
This simplifies to:
\[ x \, dy = (xy - y) \, dx \]
Factoring out \( y \) gives us:
\[ x \, dy = y(x - 1) \, dx \]

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

### Step 3: Integrating Both Sides
Next, we integrate both sides:
\[ \int \frac{dy}{y} = \int \left(1 - \frac{1}{x}\right) \, dx \]
The left side integrates to:
\[ \ln |y| + C_1 \]
The right side integrates to:
\[ x - \ln |x| + C_2 \]
Thus, we have:
\[ \ln |y| = x - \ln |x| + C \]
where \( C = C_2 - C_1 \).

### Step 4: Exponentiating Both Sides
Exponentiating both sides to eliminate the logarithm gives us:
\[ |y| = e^{x - \ln |x| + C} \]
This can be simplified to:
\[ y = \frac{e^C \cdot e^x}{|x|} \]
Let \( k = e^C \), then:
\[ y = \frac{k e^x}{x} \]

### Step 5: Applying the Initial Condition
Now we apply the initial condition \( y(1) = 1 \):
\[ 1 = \frac{k e^1}{1} \]
This simplifies to:
\[ k = \frac{1}{e} \]
Thus, substituting back, we get:
\[ y = \frac{1}{e} \cdot \frac{e^x}{x} = \frac{e^{x-1}}{x} \]

### Final Solution
The solution to the differential equation is:
\[ y = \frac{e^{x-1}}{x} \]