If any differential equation in the form
`f(f_(1)(x,y)d(f_(1)(x,y)+phi(f_(2)(x,y)d(f_(2)(x,y))+....=0`
then each term can be intergrated separately.
For example,
`intsinxyd(xy)+int((x)/(y))d((x)/(y))=-cos xy+(1)/(2)((x)/(y))^(2)+C`
Solution of differential equation
`(2xcosy+y^(2)cosx)dx+(2ysinx-x^(2)siny)dy=0` is

Options

A. `x^(2)cosy+y^(2)sinx=C`
B. `xcosy+ysinx=C`
C. `x^(2)cos^(2)y+y^(2)sin^(2)x=C`
D. None of the above

Answer

To solve the differential equation given by

\[ (2x \cos y + y^2 \cos x) dx + (2y \sin x - x^2 \sin y) dy = 0, \]

we will separate the terms and integrate them accordingly.

### Step 1: Rearranging the Equation

We start with the given equation:

\[ 2x \cos y \, dx + y^2 \cos x \, dx + (2y \sin x - x^2 \sin y) \, dy = 0. \]

We can rearrange it as:

\[ (2x \cos y + y^2 \cos x) \, dx + (2y \sin x - x^2 \sin y) \, dy = 0. \]

### Step 2: Identifying the Terms

We can identify the terms in the equation:

- Let \( M = 2x \cos y + y^2 \cos x \)
- Let \( N = 2y \sin x - x^2 \sin y \)

### Step 3: Finding the Partial Derivatives

To check if this is an exact differential equation, we compute the partial derivatives:

\[ \frac{\partial M}{\partial y} = -2x \sin y + 2y \cos x, \]
\[ \frac{\partial N}{\partial x} = 2y \cos x - 2x \sin y. \]

Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}\), the equation is not exact. However, we can still proceed with the integration.

### Step 4: Integrating Each Term

We will integrate each term separately.

1. **Integrating \( M \) with respect to \( x \)**:

\[ \int (2x \cos y + y^2 \cos x) \, dx = x^2 \cos y + y^2 \sin x + g(y), \]

where \( g(y) \) is an arbitrary function of \( y \).

2. **Integrating \( N \) with respect to \( y \)**:

\[ \int (2y \sin x - x^2 \sin y) \, dy = y^2 \sin x + x^2 \cos y + h(x), \]

where \( h(x) \) is an arbitrary function of \( x \).

### Step 5: Combining the Results

Now, we combine the results from both integrations. The general solution can be expressed as:

\[ x^2 \cos y + y^2 \sin x = C, \]

where \( C \) is a constant.

### Final Solution

Thus, the solution to the differential equation is:

\[ x^2 \cos y + y^2 \sin x = C. \]