The curve that satisfies the differential equation `y'=(x^(2) + y^2)/(2xy)` and passes through (2,1) is a hyperbola with eccentricity :

Options

A. `sqrt(2) `
B. `sqrt(3)`
C. `2`
D. `sqrt(5)`

Answer

To solve the differential equation \( y' = \frac{x^2 + y^2}{2xy} \) and find the eccentricity of the hyperbola that passes through the point (2, 1), we will follow these steps:

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

### Step 2: Substitute \( y = vx \)
Since the equation is homogeneous, we can use the substitution \( y = vx \), where \( v = \frac{y}{x} \). Then, we have:
\[ \frac{dy}{dx} = x \frac{dv}{dx} + v \]
Substituting \( y = vx \) into the differential equation gives:
\[ \frac{dy}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2 + v^2x^2}{2vx^2} = \frac{1 + v^2}{2v} \]

### Step 3: Equate and Rearrange
Now we equate the two expressions for \( \frac{dy}{dx} \):
\[ x \frac{dv}{dx} + v = \frac{1 + v^2}{2v} \]
Rearranging gives:
\[ x \frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v} \]

### Step 4: Separate Variables
Now we separate the variables:
\[ 2v \, dv = \frac{1 - v^2}{x} \, dx \]
This can be rewritten as:
\[ \frac{2v}{1 - v^2} \, dv = \frac{1}{x} \, dx \]

### Step 5: Integrate Both Sides
Integrating both sides:
\[ \int \frac{2v}{1 - v^2} \, dv = \int \frac{1}{x} \, dx \]
The left side can be integrated using substitution:
\[ \int \frac{2v}{1 - v^2} \, dv = -\ln|1 - v^2| + C_1 \]
The right side integrates to:
\[ \ln|x| + C_2 \]
Thus, we have:
\[ -\ln|1 - v^2| = \ln|x| + C \]

### Step 6: Exponentiate
Exponentiating both sides gives:
\[ |1 - v^2| = \frac{K}{|x|} \quad \text{(where \( K = e^{-C} \))} \]

### Step 7: Substitute Back for \( v \)
Substituting back \( v = \frac{y}{x} \):
\[ |1 - \left(\frac{y}{x}\right)^2| = \frac{K}{|x|} \]
This leads to the equation of the hyperbola:
\[ x^2 - y^2 = Kx \]

### Step 8: Find the Constant \( K \)
Using the point (2, 1):
\[ 2^2 - 1^2 = K \cdot 2 \implies 4 - 1 = 2K \implies 3 = 2K \implies K = \frac{3}{2} \]

### Step 9: Write the Final Equation
The equation of the hyperbola becomes:
\[ x^2 - y^2 - \frac{3}{2}x = 0 \]

### Step 10: Standard Form of the Hyperbola
Rearranging gives:
\[ x^2 - \frac{3}{2}x - y^2 = 0 \]
Completing the square:
\[ \left(x - \frac{3}{4}\right)^2 - y^2 = \frac{9}{16} \]
This is in the form:
\[ \frac{\left(x - \frac{3}{4}\right)^2}{\frac{9}{16}} - \frac{y^2}{\frac{9}{16}} = 1 \]

### Step 11: Identify \( a \) and \( b \)
From the equation, we have:
- \( a^2 = \frac{9}{16} \) so \( a = \frac{3}{4} \)
- \( b^2 = \frac{9}{16} \) so \( b = \frac{3}{4} \)

### Step 12: Calculate the Eccentricity
Using the formula for eccentricity \( e = \sqrt{1 + \frac{b^2}{a^2}} \):
\[ e = \sqrt{1 + \frac{\frac{9}{16}}{\frac{9}{16}}} = \sqrt{1 + 1} = \sqrt{2} \]

### Final Answer
The eccentricity of the hyperbola is \( \sqrt{2} \).

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