Solve `(dy)/(dx) = yf^(')(x) = f(x) f^(')(x)`, where `f(x)` is a given integrable function of `x`.

Explanation

To solve the differential equation

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

where \( f(x) \) is a given integrable function of \( x \), we can follow these steps:

### Step 1: Rearranging the equation
We start by rearranging the equation to isolate \(\frac{dy}{dx}\):

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

### Step 2: Factoring out \( f'(x) \)
Next, we can factor out \( f'(x) \) from the right-hand side:

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

### Step 3: Introducing a substitution
Let’s introduce a substitution to simplify the equation. Set \( t = f(x) - y \). Then we have:

\[ y = f(x) - t \]

Differentiating both sides with respect to \( x \):

\[ \frac{dy}{dx} = f'(x) - \frac{dt}{dx} \]

### Step 4: Substituting back into the equation
Now we substitute this back into our rearranged equation:

\[ f'(x) - \frac{dt}{dx} = f'(x) (f(x) - (f(x) - t)) \]

This simplifies to:

\[ f'(x) - \frac{dt}{dx} = f'(x) t \]

### Step 5: Rearranging the equation again
Rearranging gives us:

\[ f'(x) - f'(x) t = \frac{dt}{dx} \]

Factoring out \( f'(x) \):

\[ f'(x)(1 - t) = \frac{dt}{dx} \]

### Step 6: Separating variables
Now we can separate the variables:

\[ \frac{dt}{1 - t} = f'(x) dx \]

### Step 7: Integrating both sides
Integrate both sides:

\[ \int \frac{dt}{1 - t} = \int f'(x) dx \]

The left side integrates to:

\[ -\log|1 - t| + C_1 \]

The right side integrates to:

\[ f(x) + C_2 \]

Combining these gives:

\[ -\log|1 - t| = f(x) + C \]

where \( C = C_2 - C_1 \).

### Step 8: Solving for \( t \)
Now we can solve for \( t \):

\[ \log|1 - t| = -f(x) - C \]

Exponentiating both sides:

\[ |1 - t| = e^{-f(x) - C} = \frac{K}{e^{f(x)}} \]

where \( K = e^{-C} \) is a constant.

### Step 9: Substituting back for \( y \)
Recall that \( t = f(x) - y \), so:

\[ |1 - (f(x) - y)| = \frac{K}{e^{f(x)}} \]

This simplifies to:

\[ |y - f(x) + 1| = \frac{K}{e^{f(x)}} \]

### Final Step: Writing the solution
Thus, the general solution is:

\[ y - f(x) + 1 = \pm \frac{K}{e^{f(x)}} \]

or

\[ y = f(x) - 1 \pm \frac{K}{e^{f(x)}} \]