Solution of the differential equation `(1+x)(dy)/(dx)-xy=1-x` is

Options

A. `xy+y+x=ce^(x)`
B. `xy+y-x=ce^(x)`
C. `xy+y+x=ce^(-x)`
D. `xy+y-x=ce^(-x)`

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (1+x)\frac{dy}{dx} - xy = 1 - x \]
We can rearrange this to isolate \(\frac{dy}{dx}\):
\[ (1+x)\frac{dy}{dx} = xy + 1 - x \]
Now, divide both sides by \(1+x\):
\[ \frac{dy}{dx} = \frac{xy + 1 - x}{1+x} \]

### Step 2: Simplifying the Right Side
We can simplify the right side:
\[ \frac{dy}{dx} = \frac{xy}{1+x} + \frac{1 - x}{1+x} \]
This gives us:
\[ \frac{dy}{dx} = \frac{xy}{1+x} + \frac{1}{1+x} - \frac{x}{1+x} \]
Thus, we can rewrite it as:
\[ \frac{dy}{dx} + \frac{-x}{1+x}y = \frac{1-x}{1+x} \]

### Step 3: Identifying the Form
This is now in the standard linear form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \(P(x) = -\frac{x}{1+x}\) and \(Q(x) = \frac{1-x}{1+x}\).

### Step 4: Finding the Integrating Factor
The integrating factor \(I(x)\) is given by:
\[ I(x) = e^{\int P(x) \, dx} = e^{\int -\frac{x}{1+x} \, dx} \]
To compute this integral, we can use substitution. Let \(u = 1+x\), then \(du = dx\) and \(x = u - 1\):
\[ -\int \frac{x}{1+x} \, dx = -\int \frac{u-1}{u} \, du = -\int (1 - \frac{1}{u}) \, du = -\left(u - \ln|u|\right) + C \]
Substituting back \(u = 1+x\):
\[ -\left((1+x) - \ln|1+x|\right) + C = -1 - x + \ln|1+x| + C \]
Thus, the integrating factor is:
\[ I(x) = e^{-1-x+\ln|1+x|} = \frac{1+x}{e^{1+x}} \]

### Step 5: Solving the Differential Equation
Now we multiply the entire differential equation by the integrating factor:
\[ \frac{1+x}{e^{1+x}}\frac{dy}{dx} + \frac{-x(1+x)}{(1+x)e^{1+x}}y = \frac{(1-x)(1+x)}{(1+x)e^{1+x}} \]
This simplifies to:
\[ \frac{d}{dx}\left(y \cdot \frac{1+x}{e^{1+x}}\right) = \frac{(1-x)}{e^{1+x}} \]

### Step 6: Integrating Both Sides
Integrate both sides:
\[ \int \frac{d}{dx}\left(y \cdot \frac{1+x}{e^{1+x}}\right) \, dx = \int \frac{(1-x)}{e^{1+x}} \, dx \]
The left side simplifies to:
\[ y \cdot \frac{1+x}{e^{1+x}} = \int \frac{(1-x)}{e^{1+x}} \, dx + C \]

### Step 7: Finding the Final Solution
After integrating the right side, we can express \(y\) in terms of \(x\):
\[ y = \frac{e^{1+x}}{1+x}\left(\int \frac{(1-x)}{e^{1+x}} \, dx + C\right) \]

### Final Answer
This gives us the general solution of the differential equation.