Integrating factor of the differential equation `(x+1)(dy)/(dx)-y=e^(3x)(x+1)^(2)`, is

Options

A. `-(x+1)`
B. `log(x+1)`
C. `e^(x+1)`
D. `(1)/(x+1)`

Explanation

To find the integrating factor of the given differential equation \((x+1)\frac{dy}{dx} - y = e^{3x}(x+1)^{2}\), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the differential equation in standard linear form. We can divide the entire equation by \(x + 1\):

\[ \frac{dy}{dx} - \frac{y}{x + 1} = e^{3x}(x + 1) \]

### Step 2: Identify \(p(x)\)
In the standard form \(\frac{dy}{dx} + p(x)y = q(x)\), we identify \(p(x)\) as the coefficient of \(y\). Here, \(p(x) = -\frac{1}{x + 1}\).

### Step 3: Find the Integrating Factor
The integrating factor \(\mu(x)\) is given by the formula:

\[ \mu(x) = e^{\int p(x) \, dx} \]

Substituting \(p(x)\):

\[ \mu(x) = e^{\int -\frac{1}{x + 1} \, dx} \]

### Step 4: Calculate the Integral
Calculating the integral:

\[ \int -\frac{1}{x + 1} \, dx = -\ln|x + 1| + C \]

Thus, we have:

\[ \mu(x) = e^{-\ln|x + 1|} = \frac{1}{|x + 1|} \]

Since \(x + 1\) is positive for \(x > -1\), we can simplify this to:

\[ \mu(x) = \frac{1}{x + 1} \]

### Step 5: Conclusion
The integrating factor of the differential equation is:

\[ \mu(x) = \frac{1}{x + 1} \]