The equation of the curve satisfying the differential equation `y^2 (x^2 + 1) = 2xy` passing through the point (0,1) and having slope of tangnet at `x = 0` as 3, is (Here `y=(dy)/(dx)andy_2=(d^2y)/(dx^2))`

Options

A. `y=x^(2)+3x+2`
B. `y^(2)=x^(2)+3x+1`
C. `y=x^(3)+3x+1`
D. none of these

Answer

To solve the given differential equation \( y^2 (x^2 + 1) = 2xy \) and find the equation of the curve that passes through the point (0,1) with a slope of the tangent at \( x = 0 \) as 3, we can follow these steps:

### Step 1: Rewrite the Differential Equation
The given equation can be rewritten in terms of derivatives:
\[ y^2 (x^2 + 1) = 2xy \]
This implies:
\[ \frac{d^2y}{dx^2} (x^2 + 1) = 2x \frac{dy}{dx} \]

### Step 2: Substitute Variables
Let \( \frac{dy}{dx} = t \). Then, \( \frac{d^2y}{dx^2} = \frac{dt}{dx} \). Substituting these into the equation gives:
\[ \frac{dt}{dx} (x^2 + 1) = 2xt \]

### Step 3: Rearranging the Equation
Rearranging the equation, we have:
\[ \frac{dt}{t} = \frac{2x}{x^2 + 1} dx \]

### Step 4: Integrate Both Sides
Integrate both sides:
\[ \int \frac{dt}{t} = \int \frac{2x}{x^2 + 1} dx \]
The left side integrates to \( \ln |t| \). For the right side, we can use substitution \( u = x^2 + 1 \), \( du = 2x dx \):
\[ \int \frac{2x}{x^2 + 1} dx = \ln |x^2 + 1| + C \]
Thus, we have:
\[ \ln |t| = \ln |x^2 + 1| + C \]

### Step 5: Exponentiate Both Sides
Exponentiating both sides gives:
\[ t = k(x^2 + 1) \]
where \( k = e^C \).

### Step 6: Substitute Back for \( t \)
Substituting back for \( t \):
\[ \frac{dy}{dx} = k(x^2 + 1) \]

### Step 7: Integrate Again
Integrate again to find \( y \):
\[ y = \int k(x^2 + 1) dx = k\left(\frac{x^3}{3} + x\right) + C_1 \]

### Step 8: Apply Initial Conditions
At \( x = 0 \), \( y = 1 \):
\[ 1 = k\left(0 + 0\right) + C_1 \implies C_1 = 1 \]
Thus, we have:
\[ y = k\left(\frac{x^3}{3} + x\right) + 1 \]

### Step 9: Use the Slope Condition
We know that the slope at \( x = 0 \) is 3:
\[ \frac{dy}{dx} = k(0 + 1) = k \]
Setting \( k = 3 \):
\[ y = 3\left(\frac{x^3}{3} + x\right) + 1 = x^3 + 3x + 1 \]

### Final Solution
The equation of the curve is:
\[ \boxed{y = x^3 + 3x + 1} \]