The integrating factor of `(1+y^(2)) dx = (tan^(-1)y-x) dy` is -

Options

A. `tan^(-1) y`
B. tan y
C. `e^(tan^(-1)y)`
D. `e^(tan y)`

Answer

To find the integrating factor for the given differential equation \((1+y^2)dx = (tan^{-1}y - x)dy\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
We start by rewriting the equation in the form \(\frac{dx}{dy} + P(y)x = Q(y)\).

Given:
\[ (1+y^2)dx = (tan^{-1}y - x)dy \]
We can rearrange it:
\[ dx = \frac{(tan^{-1}y - x)}{(1+y^2)} dy \]
Now, we can express it as:
\[ \frac{dx}{dy} + \frac{x}{1+y^2} = \frac{tan^{-1}y}{1+y^2} \]

### Step 2: Identify \(P(y)\) and \(Q(y)\)
From the equation \(\frac{dx}{dy} + P(y)x = Q(y)\), we identify:
- \(P(y) = \frac{1}{1+y^2}\)
- \(Q(y) = \frac{tan^{-1}y}{1+y^2}\)

### Step 3: Find the integrating factor
The integrating factor \(\mu(y)\) is given by:
\[ \mu(y) = e^{\int P(y) dy} \]
Calculating the integral:
\[ \int P(y) dy = \int \frac{1}{1+y^2} dy = \tan^{-1}y + C \]
Thus, the integrating factor becomes:
\[ \mu(y) = e^{\tan^{-1}y} \]

### Step 4: Conclusion
The integrating factor for the given differential equation is:
\[ e^{\tan^{-1}y} \]