General solution of differential equation of `f (x) (dy)/(dx) =f ^(2) (x)+yf(x) +f'(x)y` is:
(c being arbitary constant.)

Options

A. `y=f (x) + ce ^(x)`
B. `y=-f (x )+ce ^(x)`
C. `y=-f (x) + ce ^(x)f (x)`
D. `y=c f (x) +e ^(x)`

Answer

To solve the differential equation \( f(x) \frac{dy}{dx} = f^2(x) + yf(x) + f'(x)y \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by dividing both sides by \( f(x) \) (assuming \( f(x) \neq 0 \)):
\[ \frac{dy}{dx} = f(x) + y + \frac{f'(x)}{f(x)} y \]

### Step 2: Grouping Terms
Next, we can rewrite the equation as:
\[ \frac{dy}{dx} - \left(1 + \frac{f'(x)}{f(x)}\right)y = f(x) \]
This is now in the standard form of a linear differential equation:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \( P(x) = -\left(1 + \frac{f'(x)}{f(x)}\right) \) and \( Q(x) = f(x) \).

### Step 3: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{-\int \left(1 + \frac{f'(x)}{f(x)}\right) \, dx} \]
This can be simplified as:
\[ \mu(x) = e^{-\int dx - \int \frac{f'(x)}{f(x)} \, dx} = e^{-x} \cdot \frac{1}{f(x)} \]

### Step 4: Applying the Integrating Factor
We multiply the entire differential equation by the integrating factor:
\[ e^{-x} \cdot \frac{1}{f(x)} \frac{dy}{dx} - e^{-x} \cdot \frac{1 + \frac{f'(x)}{f(x)}}{f(x)} y = e^{-x} f(x) \]

### Step 5: Simplifying
This simplifies to:
\[ \frac{d}{dx}\left(y \cdot e^{-x} \cdot \frac{1}{f(x)}\right) = e^{-x} \]

### Step 6: Integrating Both Sides
Integrating both sides with respect to \( x \):
\[ y \cdot e^{-x} \cdot \frac{1}{f(x)} = -e^{-x} + C \]
where \( C \) is the constant of integration.

### Step 7: Solving for \( y \)
Now, we can solve for \( y \):
\[ y = -f(x) + C \cdot e^{x} \cdot f(x) \]

### Final Solution
Thus, the general solution of the given differential equation is:
\[ y = -f(x) + C \cdot e^{x} f(x) \]