The equation of the curve satisfying the equation `( 1 + y^2) dx + ( x-e^(-tan^(-1)y))dy=0` and passing through origin is

Options

A. ` x e^(- tan^(-1)y)= cos^(-1)y`
B. `xe^(- cot^(-1)y) = tan^(-1)y`
C. `y tan^(-1) y = tan^(-1) x `
D. `xe^(tan^(-1)y) = tan^(-1) y`

Answer

To solve the differential equation \( (1 + y^2)dx + (x - e^{-\tan^{-1}y})dy = 0 \) and find the equation of the curve passing through the origin, we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ (1 + y^2)dx + (x - e^{-\tan^{-1}y})dy = 0 \]
We can rearrange this to:
\[ (1 + y^2)dx = - (x - e^{-\tan^{-1}y})dy \]
or
\[ dx + \frac{x - e^{-\tan^{-1}y}}{1 + y^2}dy = 0 \]

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

### Step 3: Find the Integrating Factor
The integrating factor \( IF \) for this type of equation can be found using:
\[ IF = e^{\int P(y) dy} \]
Calculating \( P(y) \):
\[ \int P(y) dy = \int \frac{1}{1 + y^2} dy = \tan^{-1}(y) \]
Thus, the integrating factor is:
\[ IF = e^{\tan^{-1}(y)} \]

### Step 4: Multiply the Equation by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor:
\[ e^{\tan^{-1}(y)}(1 + y^2)dx + e^{\tan^{-1}(y)}(x - e^{-\tan^{-1}y})dy = 0 \]

### Step 5: Solve the Equation
The solution of this type of equation can be expressed as:
\[ X \cdot IF = \int IF \cdot Q(y) dy \]
Substituting \( IF \) and \( Q(y) \):
\[ X \cdot e^{\tan^{-1}(y)} = \int e^{\tan^{-1}(y)} \left( e^{-\tan^{-1}(y)} \right) dy \]
This simplifies to:
\[ X \cdot e^{\tan^{-1}(y)} = \int \frac{dy}{1 + y^2} \]
Integrating gives:
\[ X \cdot e^{\tan^{-1}(y)} = \tan^{-1}(y) + C \]

### Step 6: Apply the Initial Condition
Since the curve passes through the origin \((0, 0)\), we substitute \( x = 0 \) and \( y = 0 \):
\[ 0 \cdot e^{\tan^{-1}(0)} = \tan^{-1}(0) + C \]
This implies:
\[ 0 = 0 + C \Rightarrow C = 0 \]

### Step 7: Final Solution
Thus, the equation of the curve is:
\[ X \cdot e^{\tan^{-1}(y)} = \tan^{-1}(y) \]
or
\[ X = \frac{\tan^{-1}(y)}{e^{\tan^{-1}(y)}} \]