IF ` x ( dy )/(dx) + y=x . ( f ( x . y) )/( f'(x.y)) ` then f ( x . y) is equal to ( K being an arbitary constant )

Options

A. `Ke ^(x^2 //2)`
B. `Ke ^(y^2//2)`
C. `Ke^(xy//2)`
D. none of these

Answer

To solve the given differential equation

\[ x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}, \]

we will follow these steps:

### Step 1: Rearranging the Equation
First, we can rearrange the equation to isolate the derivative term. We can write it as:

\[ x \frac{dy}{dx} = x \frac{f(xy)}{f'(xy)} - y. \]

### Step 2: Taking LCM
Next, we can take the LCM on the left-hand side:

\[ x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}. \]

This can be rewritten as:

\[ \frac{d(xy)}{dx} = x \frac{f(xy)}{f'(xy)}. \]

### Step 3: Using the Product Rule
Using the product rule of differentiation, we have:

\[ \frac{d(xy)}{dx} = x \frac{dy}{dx} + y. \]

### Step 4: Rearranging the Equation
Now we can rearrange the equation to:

\[ \frac{f'(xy)}{f(xy)} \frac{d(xy)}{dx} = x. \]

### Step 5: Integrating Both Sides
Now, we will integrate both sides. We can set \( t = xy \), then \( dt = y \, dx + x \, dy \). The left-hand side becomes:

\[ \int \frac{f'(t)}{f(t)} dt = \int x \, dx. \]

### Step 6: Applying Integration Result
Using the standard result of integration:

\[ \int \frac{f'(t)}{f(t)} dt = \log |f(t)| + C, \]

we can write:

\[ \log |f(t)| = \frac{x^2}{2} + C. \]

### Step 7: Exponentiating Both Sides
Exponentiating both sides gives:

\[ f(t) = e^{\frac{x^2}{2} + C} = e^{C} e^{\frac{x^2}{2}}. \]

Let \( K = e^{C} \) (where \( K \) is an arbitrary constant), so we have:

\[ f(xy) = K e^{\frac{x^2}{2}}. \]

### Final Result
Thus, the final result is:

\[ f(xy) = K e^{\frac{x^2}{2}}, \]

where \( K \) is an arbitrary constant.

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