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 \(\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}\) with the condition \(f(1) = 2\), we will follow these steps:

### Step 1: Substitute \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we can differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]

### Step 2: Substitute into the differential equation
Substituting \(y = vx\) into the differential equation gives:
\[ v + x\frac{dv}{dx} = \frac{(vx)^2 - x^2}{2x(vx)} = \frac{v^2x^2 - x^2}{2vx^2} = \frac{x^2(v^2 - 1)}{2vx^2} = \frac{v^2 - 1}{2v} \]

### Step 3: Rearranging the equation
We can rearrange this equation:
\[ x\frac{dv}{dx} = \frac{v^2 - 1}{2v} - v \]
\[ x\frac{dv}{dx} = \frac{v^2 - 1 - 2v^2}{2v} = \frac{-v^2 - 1}{2v} \]
\[ x\frac{dv}{dx} = -\frac{(v^2 + 1)}{2v} \]

### Step 4: Separate variables
Now, 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:
\[ \ln(v^2 + 1) \]
And the right side integrates to:
\[ -\ln(x) + C \]
Thus, we have:
\[ \ln(v^2 + 1) = -\ln(x) + C \]

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

### Step 7: Substitute back for \(y\)
Recall that \(v = \frac{y}{x}\), so:
\[ \frac{y^2}{x^2} + 1 = \frac{C}{x} \]
Multiplying through by \(x^2\):
\[ y^2 + x^2 = Cx \]

### Step 8: Use the initial condition to find \(C\)
Using the condition \(f(1) = 2\) (which means \(y(1) = 2\)):
\[ 2^2 + 1^2 = C \cdot 1 \implies 4 + 1 = C \implies C = 5 \]
Thus, we have:
\[ y^2 + x^2 = 5x \]

### Step 9: Differentiate to find \(f'(x)\)
Now, we differentiate \(y^2 + x^2 = 5x\):
\[ 2y\frac{dy}{dx} + 2x = 5 \]
Solving for \(\frac{dy}{dx}\):
\[ 2y\frac{dy}{dx} = 5 - 2x \implies \frac{dy}{dx} = \frac{5 - 2x}{2y} \]

### Step 10: Evaluate at \(x = 1\)
Substituting \(x = 1\) and \(y = 2\):
\[ \frac{dy}{dx} = \frac{5 - 2 \cdot 1}{2 \cdot 2} = \frac{5 - 2}{4} = \frac{3}{4} \]

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

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