The solution of `(dy)/(dx)=(x^2+y^2+1)/(2x y)` satisfying `y(1)=1` is given by

Options

A. a system of parabolas
B. a system of circles
C. `y^(2)=x(1+x)-1`
D. `(x-2)^(2)+(y-3)^(2)=5`

Explanation

To solve the differential equation \(\frac{dy}{dx} = \frac{x^2 + y^2 + 1}{2xy}\) with the initial condition \(y(1) = 1\), we will follow these steps:

### Step 1: Rewrite the Equation
We start by rewriting the given equation:
\[ 2xy \frac{dy}{dx} = x^2 + y^2 + 1 \]

### Step 2: Rearranging the Equation
Next, we rearrange the equation:
\[ 2xy \frac{dy}{dx} - y^2 = x^2 + 1 \]

### Step 3: Divide by \(x\)
Now, we divide both sides by \(x\):
\[ 2y \frac{dy}{dx} - \frac{y^2}{x} = x + \frac{1}{x} \]

### Step 4: Substitute \(y^2 = v\)
Let \(v = y^2\). Then, differentiating both sides gives:
\[ 2y \frac{dy}{dx} = \frac{dv}{dx} \]
Substituting this into our equation, we have:
\[ \frac{dv}{dx} - \frac{v}{x} = x + \frac{1}{x} \]

### Step 5: Identify \(p\) and \(q\)
This is a linear differential equation of the form \(\frac{dv}{dx} + p v = q\), where:
\[ p = -\frac{1}{x}, \quad q = x + \frac{1}{x} \]

### Step 6: Find the Integrating Factor
The integrating factor \(IF\) is given by:
\[ IF = e^{\int p \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln |x|} = \frac{1}{x} \]

### Step 7: Multiply the Equation by the Integrating Factor
Multiplying the entire equation by the integrating factor:
\[ \frac{1}{x} \cdot \frac{dv}{dx} - \frac{v}{x^2} = \left(x + \frac{1}{x}\right) \cdot \frac{1}{x} \]
This simplifies to:
\[ \frac{1}{x} \cdot \frac{dv}{dx} - \frac{v}{x^2} = 1 + \frac{1}{x^2} \]

### Step 8: Integrate Both Sides
Now we integrate both sides:
\[ \int \left(\frac{1}{x} \frac{dv}{dx} - \frac{v}{x^2}\right) dx = \int \left(1 + \frac{1}{x^2}\right) dx \]
The left side integrates to:
\[ \ln |v| = x - \frac{1}{x} + C \]

### Step 9: Solve for \(v\)
Exponentiating both sides gives:
\[ v = e^{x - \frac{1}{x} + C} = Ce^{x - \frac{1}{x}} \]
Since \(v = y^2\), we have:
\[ y^2 = Ce^{x - \frac{1}{x}} \]

### Step 10: Apply Initial Condition
Using the initial condition \(y(1) = 1\):
\[ 1^2 = Ce^{1 - 1} \implies 1 = C \implies C = 1 \]
Thus, we have:
\[ y^2 = e^{x - \frac{1}{x}} \]

### Final Answer
The final solution is:
\[ y^2 = e^{x - \frac{1}{x}} \]