Find the general solution of the following differential equations :
`x y'-y=(x+1)e^(-x)`

Answer

To solve the differential equation \( xy' - y = (x + 1)e^{-x} \), we will follow these steps:

### Step 1: Rewrite the equation
The given equation can be rewritten in standard form:
\[ xy' - y = (x + 1)e^{-x} \]
This can be expressed as:
\[ y' - \frac{y}{x} = \frac{(x + 1)e^{-x}}{x} \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
From the standard form \( y' + p(x)y = q(x) \), we identify:
- \( p(x) = -\frac{1}{x} \)
- \( q(x) = \frac{(x + 1)e^{-x}}{x} \)

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln |x|} = \frac{1}{x} \]

### Step 4: Multiply the entire equation by the integrating factor
Multiplying the entire differential equation by \( \mu(x) \):
\[ \frac{y}{x} y' - \frac{y}{x^2} = \frac{(x + 1)e^{-x}}{x^2} \]

### Step 5: Rewrite the left-hand side
The left-hand side can be rewritten as:
\[ \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{(x + 1)e^{-x}}{x^2} \]

### Step 6: Integrate both sides
Now we integrate both sides:
\[ \int \frac{d}{dx}\left(\frac{y}{x}\right) \, dx = \int \frac{(x + 1)e^{-x}}{x^2} \, dx \]
The left-hand side simplifies to:
\[ \frac{y}{x} = \int \frac{(x + 1)e^{-x}}{x^2} \, dx + C \]

### Step 7: Solve the integral on the right-hand side
To solve the integral \( \int \frac{(x + 1)e^{-x}}{x^2} \, dx \), we can use integration by parts or substitution. However, for simplicity, we will denote the integral as \( I \):
\[ I = \int \frac{(x + 1)e^{-x}}{x^2} \, dx \]

### Step 8: Substitute back to find \( y \)
After evaluating the integral \( I \), we substitute back:
\[ y = x \left( I + C \right) \]

### Step 9: Final solution
Thus, the general solution of the differential equation is:
\[ y = x \left( \int \frac{(x + 1)e^{-x}}{x^2} \, dx + C \right) \]