`x(2y-1)dx+(x^(2)+1)dy=0`

Options

A. `(x^(2)+1)(2-y)=c`
B. `(x^(2)-1)(2y-1)=c`
C. `(x^(2)+1)(2y-1)=c`
D. `(tan^(-1)x)+y^(2)-y=c`

Answer

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

### Step 1: Rearranging the Equation
We start by rewriting the equation in a more manageable form:
\[ (x^2 + 1)dy = -x(2y - 1)dx \]
Now, we can separate the variables:
\[ \frac{dy}{2y - 1} = -\frac{x}{x^2 + 1}dx \]

### Step 2: Integrating Both Sides
Next, we will integrate both sides. The left side requires a simple substitution, while the right side can be integrated directly:
\[ \int \frac{dy}{2y - 1} = \int -\frac{x}{x^2 + 1}dx \]

#### Left Side Integration
For the left side:
\[ \int \frac{dy}{2y - 1} = \frac{1}{2} \ln |2y - 1| + C_1 \]

#### Right Side Integration
For the right side, we can use the substitution \( t = x^2 + 1 \), then \( dt = 2x dx \) or \( x dx = \frac{1}{2} dt \):
\[ \int -\frac{x}{x^2 + 1}dx = -\frac{1}{2} \int \frac{1}{t}dt = -\frac{1}{2} \ln |t| + C_2 \]
Substituting back \( t = x^2 + 1 \):
\[ -\frac{1}{2} \ln |x^2 + 1| + C_2 \]

### Step 3: Equating the Integrals
Now we equate the results from both integrations:
\[ \frac{1}{2} \ln |2y - 1| = -\frac{1}{2} \ln |x^2 + 1| + C \]
where \( C = C_2 - C_1 \).

### Step 4: Simplifying the Equation
Multiply through by 2 to eliminate the fraction:
\[ \ln |2y - 1| = -\ln |x^2 + 1| + 2C \]
This can be rewritten using properties of logarithms:
\[ \ln |2y - 1| + \ln |x^2 + 1| = 2C \]
This simplifies to:
\[ \ln |(2y - 1)(x^2 + 1)| = 2C \]

### Step 5: Exponentiating Both Sides
Exponentiating both sides gives us:
\[ |(2y - 1)(x^2 + 1)| = e^{2C} \]
Let \( k = e^{2C} \), a positive constant:
\[ (2y - 1)(x^2 + 1) = k \]

### Step 6: Final Solution
Thus, the general solution of the differential equation is:
\[ 2y - 1 = \frac{k}{x^2 + 1} \]
or
\[ y = \frac{k}{2(x^2 + 1)} + \frac{1}{2} \]