Solve the differential equation
`(x+1)dy-2xydx=0`

Answer

To solve the differential equation \((x+1)dy - 2xy dx = 0\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (x + 1) dy - 2xy dx = 0 \]
Rearranging gives us:
\[ (x + 1) dy = 2xy dx \]
Now, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{2xy}{x + 1} \]

### Step 2: Separating Variables
Next, we separate the variables \(y\) and \(x\):
\[ \frac{1}{y} dy = \frac{2x}{x + 1} dx \]

### Step 3: Simplifying the Right Side
We can simplify the right side:
\[ \frac{2x}{x + 1} = 2 \left(\frac{x + 1 - 1}{x + 1}\right) = 2 \left(1 - \frac{1}{x + 1}\right) = 2 - \frac{2}{x + 1} \]
Thus, we can rewrite the equation as:
\[ \frac{1}{y} dy = \left(2 - \frac{2}{x + 1}\right) dx \]

### Step 4: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{1}{y} dy = \int \left(2 - \frac{2}{x + 1}\right) dx \]
The left side integrates to:
\[ \log |y| + C_1 \]
The right side integrates to:
\[ 2x - 2 \log |x + 1| + C_2 \]

### Step 5: Combining Constants
Combining the constants of integration, we can write:
\[ \log |y| = 2x - 2 \log |x + 1| + C \]
where \(C = C_2 - C_1\).

### Step 6: Exponentiating Both Sides
To solve for \(y\), we exponentiate both sides:
\[ |y| = e^{2x - 2 \log |x + 1| + C} = e^C \cdot e^{2x} \cdot e^{-2 \log |x + 1|} = e^C \cdot e^{2x} \cdot \frac{1}{(x + 1)^2} \]
Let \(k = e^C\), then:
\[ y = \frac{k e^{2x}}{(x + 1)^2} \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = \frac{k e^{2x}}{(x + 1)^2} \]