Equation of the curve through the point (1, 0) which satisfies the differential equation `(1+y^(2))dx-xy dy =0 ` is

Options

A. `x^(2)+y^(2)=1`
B. `x^(2)-y^(2)=1`
C. `2x^(2)+y^(2)=2`
D. None of these

Answer

To solve the differential equation \((1 + y^2)dx - xy dy = 0\) and find the equation of the curve that passes through the point (1, 0), we will follow these steps:

### Step 1: Rearranging the Differential Equation
We start with the given differential equation:
\[ (1 + y^2)dx - xy dy = 0 \]
Rearranging gives us:
\[ (1 + y^2)dx = xy dy \]
Now, we can separate the variables:
\[ \frac{dx}{xy} = \frac{dy}{1 + y^2} \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. The left side becomes:
\[ \int \frac{dx}{x} = \ln |x| + C_1 \]
And the right side becomes:
\[ \int \frac{dy}{1 + y^2} = \tan^{-1}(y) + C_2 \]
Thus, we have:
\[ \ln |x| = \tan^{-1}(y) + C \]
where \(C = C_2 - C_1\) is a constant.

### Step 3: Exponentiating to Eliminate the Logarithm
To eliminate the logarithm, we exponentiate both sides:
\[ |x| = e^{\tan^{-1}(y) + C} = e^C \cdot e^{\tan^{-1}(y)} \]
Let \(k = e^C\), so we can write:
\[ x = k \cdot e^{\tan^{-1}(y)} \]

### Step 4: Finding the Constant \(k\)
We need to find the constant \(k\) using the point (1, 0):
\[ 1 = k \cdot e^{\tan^{-1}(0)} \]
Since \(\tan^{-1}(0) = 0\), we have:
\[ 1 = k \cdot e^0 \implies k = 1 \]
Thus, the equation simplifies to:
\[ x = e^{\tan^{-1}(y)} \]

### Step 5: Squaring Both Sides
To eliminate the exponential, we can square both sides:
\[ x^2 = e^{2 \tan^{-1}(y)} \]
Using the identity \(e^{2 \tan^{-1}(y)} = \frac{1 + y^2}{1 - y^2}\), we can rewrite this as:
\[ x^2 = 1 + y^2 \]

### Step 6: Rearranging the Equation
Rearranging gives us:
\[ x^2 - y^2 = 1 \]
This is the equation of the hyperbola.

### Final Answer
The equation of the curve is:
\[ x^2 - y^2 = 1 \]