Solve each of the following differential equations
`y-x(dy)/(dx)=2(y^2+(dy)/(dx))`.

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ y - x \frac{dy}{dx} = 2(y^2 + \frac{dy}{dx}) \]
We can rearrange this to group the terms involving \( \frac{dy}{dx} \):
\[ y - 2y^2 = x \frac{dy}{dx} + 2 \frac{dy}{dx} \]
This simplifies to:
\[ y - 2y^2 = (x + 2) \frac{dy}{dx} \]

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

### Step 3: Integrating Both Sides
Next, we will integrate both sides. The left side requires partial fraction decomposition:
\[ \frac{1}{y(1 - 2y)} = \frac{A}{y} + \frac{B}{1 - 2y} \]
Multiplying through by the denominator \( y(1 - 2y) \):
\[ 1 = A(1 - 2y) + By \]
Setting \( y = 0 \) gives \( A = 1 \). Setting \( y = \frac{1}{2} \) gives \( B = 2 \). Therefore:
\[ \frac{1}{y(1 - 2y)} = \frac{1}{y} + \frac{2}{1 - 2y} \]
Now we can integrate:
\[ \int \left( \frac{1}{y} + \frac{2}{1 - 2y} \right) dy = \int \frac{dx}{x + 2} \]
This leads to:
\[ \ln |y| - \ln |1 - 2y| = \ln |x + 2| + C \]

### Step 4: Simplifying the Result
Using properties of logarithms, we can combine the left side:
\[ \ln \left| \frac{y}{1 - 2y} \right| = \ln |x + 2| + C \]
Exponentiating both sides gives:
\[ \frac{y}{1 - 2y} = k(x + 2) \]
where \( k = e^C \).

### Step 5: Solving for \( y \)
Now, we can solve for \( y \):
\[ y = k(x + 2)(1 - 2y) \]
Rearranging gives:
\[ y + 2ky = k(x + 2) \implies y(1 + 2k) = k(x + 2) \]
Thus,
\[ y = \frac{k(x + 2)}{1 + 2k} \]

### Final Solution
The solution to the differential equation is:
\[ y = \frac{k(x + 2)}{1 + 2k} \]
where \( k \) is a constant.

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