Solve the following initial value problems and find the corresponding solution curves :
`x dy + y dx= xy dx, y(1)=1`

Answer

To solve the initial value problem given by the equation \( x \, dy + y \, dx = xy \, dx \) with the initial condition \( y(1) = 1 \), we will 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 \( dy \):
\[ x \, dy = xy \, dx - y \, dx \]
Factoring out \( y \, dx \) from the right side gives us:
\[ x \, dy = y(x - 1) \, dx \]

### Step 2: Dividing by \( x \)
Next, we divide both sides by \( x \):
\[ dy = \frac{y(x - 1)}{x} \, dx \]

### Step 3: Separating Variables
Now we can separate the variables \( y \) and \( x \):
\[ \frac{dy}{y} = \frac{(x - 1)}{x} \, dx \]

### Step 4: Integrating Both Sides
We integrate both sides:
\[ \int \frac{dy}{y} = \int \frac{(x - 1)}{x} \, dx \]
The left side integrates to:
\[ \ln |y| = \int \left(1 - \frac{1}{x}\right) \, dx \]
The right side integrates to:
\[ \ln |y| = x - \ln |x| + C \]

### Step 5: Simplifying the Equation
We can rewrite the equation as:
\[ \ln |y| + \ln |x| = x + C \]
This simplifies to:
\[ \ln |yx| = x + C \]

### Step 6: Exponentiating Both Sides
Exponentiating both sides gives us:
\[ yx = e^{x + C} \]
Let \( e^C = k \), where \( k \) is a constant:
\[ yx = ke^x \]

### Step 7: Solving for \( y \)
Now we can solve for \( y \):
\[ y = \frac{ke^x}{x} \]

### Step 8: Applying the Initial Condition
We apply the initial condition \( y(1) = 1 \):
\[ 1 = \frac{ke^1}{1} \implies k = \frac{1}{e} \]

### Step 9: Final Solution
Substituting \( k \) back into the equation gives us the final solution:
\[ y = \frac{1}{e} \cdot \frac{e^x}{x} = \frac{e^{x-1}}{x} \]

### Summary
The solution to the initial value problem is:
\[ y = \frac{e^{x-1}}{x} \]