The solution of `dy/dx = (x^2+y^2+1)/(2xy)` satisfying `y(1)=0` is given by

A. a hyperbola
B. a circle
C. ` y^2 = (1+x ) -10`
D. None

Options

A. a hyperbola
B. a circle
C. ` y^2 = (1+x ) -10`
D. None

Answer

To solve the differential equation given by

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

with the initial condition \(y(1) = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by multiplying both sides of the equation by \(2xy\):

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

### Step 2: Substituting Variables
Next, we can substitute \(y^2 = t\). Then, differentiating gives us:

\[ 2y \frac{dy}{dx} = \frac{dt}{dx} \]

Substituting this into our equation gives:

\[ \frac{dt}{dx} = x^2 + t + 1 \]

### Step 3: Rearranging to Linear Form
Now we rearrange the equation to the standard linear form:

\[ \frac{dt}{dx} - t = x^2 + 1 \]

Here, \(p(x) = -1\) and \(q(x) = x^2 + 1\).

### Step 4: Finding the Integrating Factor
The integrating factor \(I(x)\) is given by:

\[ I(x) = e^{\int p(x) \, dx} = e^{\int -1 \, dx} = e^{-x} \]

### Step 5: Multiplying by the Integrating Factor
Multiplying the entire differential equation by the integrating factor:

\[ e^{-x} \frac{dt}{dx} - e^{-x} t = e^{-x}(x^2 + 1) \]

### Step 6: Integrating Both Sides
The left side can be rewritten as:

\[ \frac{d}{dx}(e^{-x} t) = e^{-x}(x^2 + 1) \]

Integrating both sides gives:

\[ e^{-x} t = \int e^{-x}(x^2 + 1) \, dx \]

### Step 7: Solving the Integral
To solve the integral, we can use integration by parts or look up the integral:

\[ \int e^{-x}(x^2 + 1) \, dx \]

This integral can be computed, and we find:

\[ \int e^{-x}(x^2 + 1) \, dx = -e^{-x}(x^2 + 2x + 2) + C \]

### Step 8: Substituting Back for \(t\)
Now substituting back for \(t\):

\[ e^{-x} t = -e^{-x}(x^2 + 2x + 2) + C \]

Multiplying through by \(e^{x}\):

\[ t = -(x^2 + 2x + 2) + Ce^{x} \]

Since \(t = y^2\), we have:

\[ y^2 = -x^2 - 2x - 2 + Ce^{x} \]

### Step 9: Applying the Initial Condition
Using the initial condition \(y(1) = 0\):

\[ 0 = -1 - 2 - 2 + Ce^{1} \]

This simplifies to:

\[ 0 = -5 + Ce \implies Ce = 5 \implies C = \frac{5}{e} \]

### Final Step: Final Equation
Substituting \(C\) back into the equation gives:

\[ y^2 = -x^2 - 2x - 2 + \frac{5}{e} e^{x} \]

This is the final form of the solution.

### Conclusion
The solution describes a hyperbola, which corresponds to option A.

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