The solution of the differential ` (dy)/(dx) = (y phi '(x) -y^(2))/(phi(x))` is

Options

A. ` y= x/(phi (x) +c)`
B. `y = (phi (x))/x +c`
C. ` y = (phi(x) +c)/x `
D. `y= (phi(x))/(x +c)`

Explanation

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

### Step 1: Rewrite the Equation
Start by rewriting the given equation:
\[ \frac{dy}{dx} = \frac{y \phi'(x) - y^2}{\phi(x)} \]
This can be rearranged as:
\[ \frac{dy}{dx} = \frac{y \phi'(x)}{\phi(x)} - \frac{y^2}{\phi(x)} \]

### Step 2: Separate Variables
We can separate the variables by dividing through by \( y^2 \):
\[ \frac{1}{y^2} \frac{dy}{dx} = \frac{\phi'(x)}{\phi(x)} \cdot \frac{1}{y} - \frac{1}{\phi(x)} \]
Rearranging gives:
\[ \frac{1}{y^2} \frac{dy}{dx} + \frac{1}{\phi(x)} = \frac{\phi'(x)}{\phi(x)} \cdot \frac{1}{y} \]

### Step 3: Introduce a Substitution
Let \( t = -\frac{1}{y} \). Then, differentiating gives:
\[ \frac{dt}{dx} = \frac{1}{y^2} \frac{dy}{dx} \]
Substituting this into the equation gives:
\[ \frac{dt}{dx} + t \cdot \frac{\phi'(x)}{\phi(x)} = -\frac{1}{\phi(x)} \]

### Step 4: Identify the Integrating Factor
The equation is now in the standard form \( \frac{dt}{dx} + P(x)t = Q(x) \), where:
- \( P(x) = \frac{\phi'(x)}{\phi(x)} \)
- \( Q(x) = -\frac{1}{\phi(x)} \)

The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{\phi'(x)}{\phi(x)} \, dx} = e^{\ln(\phi(x))} = \phi(x) \]

### Step 5: Multiply the Equation by the Integrating Factor
Multiply the entire equation by \( \phi(x) \):
\[ \phi(x) \frac{dt}{dx} + t \phi'(x) = -1 \]

### Step 6: Integrate Both Sides
Now, we can integrate both sides:
\[ \int \left( \phi(x) \frac{dt}{dx} + t \phi'(x) \right) dx = \int -1 \, dx \]
The left side simplifies to:
\[ \phi(x) t = -x + C \]
where \( C \) is the constant of integration.

### Step 7: Substitute Back for \( t \)
Recall that \( t = -\frac{1}{y} \):
\[ \phi(x) \left(-\frac{1}{y}\right) = -x + C \]
This simplifies to:
\[ -\frac{\phi(x)}{y} = -x + C \]

### Step 8: Solve for \( y \)
Rearranging gives:
\[ \frac{\phi(x)}{y} = x + C \]
Thus,
\[ y = \frac{\phi(x)}{x + C} \]

### Final Solution
The solution to the differential equation is:
\[ y = \frac{\phi(x)}{x + C} \]