General solution of `y-x(dy)/(dx)=8(y^(2)+(dy)/(dx))` is

Options

A. `(x+8)(8y+1)=cy`
B. `(x+8)(8y-1)=cy`
C. `(8x+1)(y-1)=cy`
D. None of these

Answer

To solve the differential equation \( y - x \frac{dy}{dx} = 8(y^2 + \frac{dy}{dx}) \), we will follow these steps:

### Step 1: Rearrange the Equation
Start by rearranging the given equation:
\[ y - x \frac{dy}{dx} = 8y^2 + 8 \frac{dy}{dx} \]
Rearranging gives:
\[ y - 8y^2 = x \frac{dy}{dx} + 8 \frac{dy}{dx} \]
This can be simplified to:
\[ y - 8y^2 = (x + 8) \frac{dy}{dx} \]

### Step 2: Separate Variables
Now, we can express this in a form suitable for separation of variables:
\[ \frac{dy}{dx} = \frac{y - 8y^2}{x + 8} \]
This allows us to separate variables:
\[ \frac{dy}{y - 8y^2} = \frac{dx}{x + 8} \]

### Step 3: Integrate Both Sides
Next, we integrate both sides. The left side requires partial fraction decomposition:
\[ \frac{1}{y(1 - 8y)} = \frac{A}{y} + \frac{B}{1 - 8y} \]
Multiplying through by the denominator \( y(1 - 8y) \) and solving for \( A \) and \( B \) gives:
\[ 1 = A(1 - 8y) + By \]
Setting \( y = 0 \) gives \( A = 1 \). Setting \( y = \frac{1}{8} \) gives \( B = 8 \). Thus:
\[ \frac{1}{y(1 - 8y)} = \frac{1}{y} + \frac{8}{1 - 8y} \]
Now we integrate:
\[ \int \left( \frac{1}{y} + \frac{8}{1 - 8y} \right) dy = \int \frac{dx}{x + 8} \]
This results in:
\[ \ln |y| - \ln |1 - 8y| = \ln |x + 8| + C \]

### Step 4: Simplify the Result
Using properties of logarithms, we can combine the left side:
\[ \ln \left| \frac{y}{1 - 8y} \right| = \ln |x + 8| + C \]
Exponentiating both sides gives:
\[ \frac{y}{1 - 8y} = k(x + 8) \quad \text{where } k = e^C \]
Rearranging this leads to:
\[ y = \frac{k(x + 8)}{1 + 8k(x + 8)} \]

### Final General Solution
Thus, the general solution of the differential equation is:
\[ y = \frac{k(x + 8)}{1 + 8k(x + 8)} \]