Solution of the diff. eqn. `(1+x) (dy)/(dx) -xy =1-x` is ……….

Answer

To solve the differential equation \((1+x) \frac{dy}{dx} - xy = 1-x\), we will follow these steps:

### Step 1: Rearranging the Equation
First, we need to rearrange the equation to isolate \(\frac{dy}{dx}\). We do this by dividing the entire equation by \(1+x\):

\[ \frac{dy}{dx} - \frac{xy}{1+x} = \frac{1-x}{1+x} \]

### Step 2: Identifying p and q
Now, we can identify \(p\) and \(q\) from the standard form of a linear differential equation:

\[ \frac{dy}{dx} + p y = q \]

Here, we have:

\[ p = -\frac{x}{1+x}, \quad q = \frac{1-x}{1+x} \]

### Step 3: Finding the Integrating Factor
The integrating factor \(I\) is given by:

\[ I = e^{\int p \, dx} = e^{\int -\frac{x}{1+x} \, dx} \]

To solve the integral, we can use substitution. Let \(u = 1 + x\), then \(du = dx\) and \(x = u - 1\):

\[ I = e^{-\int \frac{u-1}{u} \, du} = e^{-\int (1 - \frac{1}{u}) \, du} = e^{-(u - \ln |u|)} = e^{-1-x + \ln |1+x|} = \frac{1+x}{e^{1+x}} \]

### Step 4: Multiplying the Equation by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor:

\[ \frac{1+x}{e^{1+x}} \left( \frac{dy}{dx} - \frac{xy}{1+x} \right) = \frac{1+x}{e^{1+x}} \cdot \frac{1-x}{1+x} \]

This simplifies to:

\[ \frac{1+x}{e^{1+x}} \frac{dy}{dx} - \frac{xy}{e^{1+x}} = \frac{1-x}{e^{1+x}} \]

### Step 5: Integrating Both Sides
The left side can be recognized as the derivative of a product:

\[ \frac{d}{dx} \left( y \cdot \frac{1+x}{e^{1+x}} \right) = \frac{1-x}{e^{1+x}} \]

Now, we integrate both sides:

\[ \int \frac{d}{dx} \left( y \cdot \frac{1+x}{e^{1+x}} \right) \, dx = \int \frac{1-x}{e^{1+x}} \, dx \]

### Step 6: Solving the Integral
The left side integrates to:

\[ y \cdot \frac{1+x}{e^{1+x}} = \int \frac{1-x}{e^{1+x}} \, dx + C \]

The right side can be solved using integration by parts or a suitable substitution.

### Step 7: Solving for y
After integrating and simplifying, we can isolate \(y\):

\[ y = \frac{e^{1+x}}{1+x} \left( \text{integral result} + C \right) \]

### Final Solution
Thus, the solution to the differential equation is given by:

\[ y = \frac{e^{1+x}}{1+x} \left( \int \frac{1-x}{e^{1+x}} \, dx + C \right) \]