Solution of the differential equation
`(1+y^(2))dx =(tan^(-1)y-x)dy` is

Options

A. `x e^(tan^(-1)y) =(1-tan^(-1)y)e^(tan^(-1)y)+c`
B. `x e^(tan^(-1)y)=(tan^(-1)y -1) e^(tan^(-1)y)+c`
C. `x=tan^(-1)y-1+c e^(-tan^(-1)y)`
D. None of these

Answer

To solve the differential equation \((1+y^2)dx = (\tan^{-1}y - x)dy\), we will follow the steps outlined below.

### Step 1: Rearranging the Equation
We start by rearranging the equation into a standard form. We can write it as:

\[ dx + \frac{x}{1+y^2} dy = \frac{\tan^{-1}y}{1+y^2} dy \]

### Step 2: Identifying the Linear Form
This can be recognized as a linear first-order differential equation in the form:

\[ \frac{dx}{dy} + P(y)x = Q(y) \]

where \(P(y) = \frac{1}{1+y^2}\) and \(Q(y) = \frac{\tan^{-1}y}{1+y^2}\).

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

\[ I(y) = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1}y} \]

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

\[ e^{\tan^{-1}y} dx + \frac{x e^{\tan^{-1}y}}{1+y^2} dy = \frac{\tan^{-1}y e^{\tan^{-1}y}}{1+y^2} dy \]

### Step 5: Recognizing the Left Side as a Derivative
The left-hand side can be recognized as the derivative of a product:

\[ \frac{d}{dy}(x e^{\tan^{-1}y}) = \frac{\tan^{-1}y e^{\tan^{-1}y}}{1+y^2} \]

### Step 6: Integrating Both Sides
Now we integrate both sides with respect to \(y\):

\[ \int \frac{d}{dy}(x e^{\tan^{-1}y}) dy = \int \frac{\tan^{-1}y e^{\tan^{-1}y}}{1+y^2} dy \]

This gives:

\[ x e^{\tan^{-1}y} = \int \frac{\tan^{-1}y e^{\tan^{-1}y}}{1+y^2} dy + C \]

### Step 7: Solving the Integral
To solve the integral on the right side, we can use integration by parts, where we let:

- \(u = \tan^{-1}y\) and \(dv = e^{\tan^{-1}y} \frac{1}{1+y^2} dy\)

After performing integration by parts, we find:

\[ \int \tan^{-1}y e^{\tan^{-1}y} dy = \tan^{-1}y e^{\tan^{-1}y} - \int e^{\tan^{-1}y} dy \]

### Step 8: Final Form of the Solution
After integrating and rearranging, we can express \(x\) in terms of \(y\):

\[ x = \frac{\tan^{-1}y e^{\tan^{-1}y}}{e^{\tan^{-1}y}} + C e^{-\tan^{-1}y} \]

### Conclusion
Thus, the solution of the differential equation is:

\[ x e^{\tan^{-1}y} = \tan^{-1}y + C \]