Find the particular solution of the differential equation `(xe^(y//x)+y)dx=xdy`, given that `y(1)=0`.

Answer

To find the particular solution of the differential equation \( (xe^{\frac{y}{x}} + y)dx = xdy \) given that \( y(1) = 0 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ (xe^{\frac{y}{x}} + y)dx = xdy \]
We can rearrange this to separate the variables:
\[ xe^{\frac{y}{x}} + y = x \frac{dy}{dx} \]

### Step 2: Divide by \( x \)
Next, we divide both sides by \( x \):
\[ e^{\frac{y}{x}} + \frac{y}{x} = \frac{dy}{dx} \]

### Step 3: Substitute \( v = \frac{y}{x} \)
Let \( v = \frac{y}{x} \), which implies \( y = vx \). Now, we differentiate \( y \):
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Substituting \( y \) and \( \frac{dy}{dx} \) into our equation gives:
\[ e^v + v = v + x\frac{dv}{dx} \]
The \( v \) terms cancel out:
\[ e^v = x\frac{dv}{dx} \]

### Step 4: Separate Variables
We can now separate the variables:
\[ \frac{dv}{e^v} = \frac{dx}{x} \]

### Step 5: Integrate Both Sides
Integrating both sides:
\[ \int \frac{dv}{e^v} = \int \frac{dx}{x} \]
The left side integrates to:
\[ -e^{-v} = \ln |x| + C \]

### Step 6: Solve for \( v \)
Rearranging gives:
\[ -e^{-v} = \ln |x| + C \implies e^{-v} = -(\ln |x| + C) \]
Taking the reciprocal:
\[ e^v = \frac{1}{-(\ln |x| + C)} \]

### Step 7: Substitute Back for \( y \)
Recall that \( v = \frac{y}{x} \):
\[ e^{\frac{y}{x}} = \frac{1}{-(\ln |x| + C)} \]

### Step 8: Find the Particular Solution
Now we use the initial condition \( y(1) = 0 \):
\[ e^{\frac{0}{1}} = \frac{1}{-(\ln |1| + C)} \implies 1 = \frac{1}{-0 - C} \implies C = -1 \]

### Step 9: Substitute \( C \) Back
Substituting \( C = -1 \) back into the equation:
\[ e^{\frac{y}{x}} = \frac{1}{-(\ln |x| - 1)} \implies e^{\frac{y}{x}} = \frac{1}{1 - \ln |x|} \]

### Final Step: Rearranging
Thus, the particular solution is:
\[ \ln |x| + e^{\frac{y}{x}} = 1 \]