The solution of `ydx-xdy+(1+x^(2))dx+x^(2)siny dy=0,` is given by

Options

A. `2x-y^(2)+cosy+c=0`
B. `y+1-x^(2)+xcosy+c=0`
C. `(x)/(y)+(1)/(y)-y+cosy+c=0`
D. `(y)/(x)+(1)/(x)-x+cosy+c=0`

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ y \, dx - x \, dy + (1 + x^2) \, dx + x^2 \sin y \, dy = 0 \]
We can rearrange it as:
\[ (y + 1 + x^2) \, dx + (-x + x^2 \sin y) \, dy = 0 \]

### Step 2: Dividing by \( x^2 \)
Next, we divide the entire equation by \( x^2 \):
\[ \frac{y + 1}{x^2} \, dx + \left(-\frac{x}{x^2} + \sin y\right) \, dy = 0 \]
This simplifies to:
\[ \frac{y + 1}{x^2} \, dx - \frac{1}{x} \, dy + \sin y \, dy = 0 \]

### Step 3: Identifying the Exact Form
We can express this in a more recognizable form:
\[ \frac{y + 1}{x^2} \, dx + \left(\sin y - \frac{1}{x}\right) \, dy = 0 \]
This suggests that we can use the method of exact equations.

### Step 4: Finding the Integrating Factor
To check if the equation is exact, we need to find \( M = \frac{y + 1}{x^2} \) and \( N = \sin y - \frac{1}{x} \). We compute:
\[ \frac{\partial M}{\partial y} = \frac{1}{x^2}, \quad \frac{\partial N}{\partial x} = \frac{1}{x^2} \]
Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact.

### Step 5: Integrating to Find the Solution
Now we integrate \( M \) with respect to \( x \):
\[ \int M \, dx = \int \frac{y + 1}{x^2} \, dx = -\frac{y + 1}{x} + C(y) \]
Next, we differentiate this with respect to \( y \) and set it equal to \( N \):
\[ \frac{\partial}{\partial y} \left(-\frac{y + 1}{x} + C(y)\right) = -\frac{1}{x} + C'(y) = \sin y - \frac{1}{x} \]
This gives us:
\[ C'(y) = \sin y \]
Integrating \( C'(y) \):
\[ C(y) = -\cos y + C_0 \]

### Step 6: Final Solution
Combining our results, we have:
\[ -\frac{y + 1}{x} - \cos y + C_0 = 0 \]
Rearranging gives:
\[ \frac{y + 1}{x} + \cos y + C = 0 \]
where \( C = -C_0 \).

### Conclusion
Thus, the solution to the differential equation is:
\[ \frac{y + 1}{x} + \cos y + C = 0 \]