The solution of the differential equation `y-x(dy)/(dx)=a(y^2+(dy)/(dx))` is

Options

A. `x=c (1+ay)(x+a)`
B. `y = c (1+ ay) (x-a)`
C. `ay = c (1- ay) (x+a)`
D. `y = c (1- ay) (ax+1)`

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ y - x \frac{dy}{dx} = a \left( y^2 + \frac{dy}{dx} \right) \]
Rearranging this gives:
\[ y - ay^2 = a \frac{dy}{dx} + x \frac{dy}{dx} \]
This can be expressed as:
\[ y - ay^2 = (a + x) \frac{dy}{dx} \]

### Step 2: Separating Variables
Now we can separate the variables by rewriting the equation:
\[ \frac{dy}{y - ay^2} = \frac{dx}{a + x} \]

### Step 3: Integrating Both Sides
Next, we integrate both sides. Starting with the left side:
\[ \int \frac{dy}{y(1 - ay)} = \int \frac{dx}{a + x} \]
To integrate the left side, we can use partial fractions:
\[ \frac{1}{y(1 - ay)} = \frac{A}{y} + \frac{B}{1 - ay} \]
Multiplying through by the denominator \( y(1 - ay) \) gives:
\[ 1 = A(1 - ay) + By \]
Setting \( y = 0 \) gives \( A = 1 \). Setting \( y = \frac{1}{a} \) gives \( B = a \). Thus:
\[ \frac{1}{y(1 - ay)} = \frac{1}{y} + \frac{a}{1 - ay} \]
So we can rewrite the integral as:
\[ \int \left( \frac{1}{y} + \frac{a}{1 - ay} \right) dy = \int \frac{dx}{a + x} \]
Integrating both sides:
\[ \ln |y| - \frac{1}{a} \ln |1 - ay| = \ln |a + x| + C \]

### Step 4: Simplifying the Result
We can simplify the left side:
\[ \ln \left( \frac{|y|}{|1 - ay|^{1/a}} \right) = \ln |a + x| + C \]
Exponentiating both sides gives:
\[ \frac{|y|}{|1 - ay|^{1/a}} = K |a + x| \]
where \( K = e^C \).

### Step 5: Final Form
Rearranging gives us:
\[ |y| = K |a + x| |1 - ay|^{1/a} \]
This represents the general solution of the differential equation.