Let `y=f(x)` be a solution of the differential equation `(dy)/(dx)=(y^(2)-x^(2))/(2xy)(AA x, y gt 0)`. If `f(1)=2`, then `f'(1)` is equal to

Options

A. 2
B. `(5)/(2)`
C. `(5)/(4)`
D. `(3)/(4)`

Answer

To solve the given differential equation and find \( f'(1) \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
The given differential equation is:
\[ \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \]
We can rewrite it as:
\[ \frac{dy}{dx} = \frac{y^2}{2xy} - \frac{x^2}{2xy} = \frac{y}{2x} - \frac{x}{2y} \]

### Step 2: Substitute \( y = vx \)
Let \( y = vx \), where \( v \) is a function of \( x \). Then, we have:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting this into the differential equation gives:
\[ v + x \frac{dv}{dx} = \frac{vx^2}{2x} - \frac{x}{2(vx)} = \frac{v}{2} - \frac{1}{2v} \]

### Step 3: Rearranging the Equation
Rearranging the equation, we get:
\[ x \frac{dv}{dx} = \frac{v}{2} - v - \frac{1}{2v} \]
This simplifies to:
\[ x \frac{dv}{dx} = -\frac{v}{2} - \frac{1}{2v} \]

### Step 4: Separate Variables
We can separate the variables:
\[ \frac{2v}{v^2 + 1} dv = -\frac{dx}{x} \]

### Step 5: Integrate Both Sides
Integrating both sides:
\[ \int \frac{2v}{v^2 + 1} dv = \int -\frac{dx}{x} \]
The left side integrates to:
\[ \log(v^2 + 1) \]
And the right side integrates to:
\[ -\log(x) + C \]
Thus, we have:
\[ \log(v^2 + 1) = -\log(x) + C \]

### Step 6: Exponentiate to Remove Logarithms
Exponentiating both sides gives:
\[ v^2 + 1 = \frac{C}{x} \]

### Step 7: Substitute Back for \( y \)
Recalling that \( v = \frac{y}{x} \), we have:
\[ \left(\frac{y}{x}\right)^2 + 1 = \frac{C}{x} \]
This simplifies to:
\[ y^2 + x^2 = Cx \]

### Step 8: Use Initial Condition to Find \( C \)
Given \( f(1) = 2 \):
\[ 2^2 + 1^2 = C \cdot 1 \implies 4 + 1 = C \implies C = 5 \]

### Step 9: Final Form of the Solution
Thus, the solution to the differential equation is:
\[ y^2 + x^2 = 5x \]

### Step 10: Differentiate to Find \( f'(x) \)
Differentiating both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} + 2x = 5 \]
Solving for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{5 - 2x}{2y} \]

### Step 11: Evaluate at \( x = 1 \)
Substituting \( x = 1 \) and \( y = 2 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{5 - 2 \cdot 1}{2 \cdot 2} = \frac{3}{4} \]

Thus, \( f'(1) = \frac{3}{4} \).

### Final Answer
\[ f'(1) = \frac{3}{4} \]