The equation of the curve satisfying the differential equation `xe^(x)sin ydy-(x+1)e^(x) cos ydx=ydy` and passing through the origin is

Options

A. `xe^(x)=y^(2)cosy`
B. `2xe^(x)=ycosy`
C. `2xe^(x)cosy+y^(2)=0`
D. `2xe^(x)cosy=y^(2)`

Answer

To solve the given differential equation \( xe^{x} \sin y \, dy - (x + 1)e^{x} \cos y \, dx = y \, dy \) and find the equation of the curve passing through the origin, we will follow these steps:

### Step 1: Rearranging the Differential Equation
We start by rearranging the given differential equation into a standard form:
\[ -(x + 1)e^{x} \cos y \, dx + (xe^{x} \sin y - y) \, dy = 0 \]
This can be rewritten as:
\[ M(x, y) \, dx + N(x, y) \, dy = 0 \]
where \( M(x, y) = -(x + 1)e^{x} \cos y \) and \( N(x, y) = xe^{x} \sin y - y \).

### Step 2: Checking for Exactness
To check if the equation is exact, we need to compute the partial derivatives:
\[ \frac{\partial M}{\partial y} = (x + 1)e^{x} \sin y \]
\[ \frac{\partial N}{\partial x} = e^{x} \sin y + xe^{x} \sin y \]
Now we check if \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \):
\[ \frac{\partial N}{\partial x} = e^{x} \sin y + xe^{x} \sin y = (x + 1)e^{x} \sin y \]
Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact.

### Step 3: Finding the Potential Function
Next, we find the potential function \( \psi(x, y) \) such that:
\[ \frac{\partial \psi}{\partial x} = M \quad \text{and} \quad \frac{\partial \psi}{\partial y} = N \]
Integrating \( M \) with respect to \( x \):
\[ \psi(x, y) = \int -(x + 1)e^{x} \cos y \, dx = -\cos y \int (x + 1)e^{x} \, dx \]
Using integration by parts:
\[ \int (x + 1)e^{x} \, dx = e^{x}(x + 1) - e^{x} + C = e^{x}x \]
Thus,
\[ \psi(x, y) = -\cos y \cdot e^{x}(x + 1) + g(y) \]
where \( g(y) \) is an arbitrary function of \( y \).

Next, we differentiate \( \psi \) with respect to \( y \) and set it equal to \( N \):
\[ \frac{\partial \psi}{\partial y} = -e^{x}(x + 1)(-\sin y) + g'(y) = e^{x}(x + 1) \sin y + g'(y) \]
Setting this equal to \( N \):
\[ e^{x}(x + 1) \sin y + g'(y) = xe^{x} \sin y - y \]
From this, we can find \( g'(y) \):
\[ g'(y) = -y \quad \Rightarrow \quad g(y) = -\frac{y^2}{2} + C \]

### Step 4: Forming the Complete Potential Function
Combining everything, we have:
\[ \psi(x, y) = -\cos y \cdot e^{x}(x + 1) - \frac{y^2}{2} = C \]

### Step 5: Applying the Initial Condition
Since the curve passes through the origin \( (0, 0) \):
\[ -\cos(0) \cdot e^{0}(0 + 1) - \frac{0^2}{2} = C \quad \Rightarrow \quad -1 = C \]
Thus, the equation becomes:
\[ -\cos y \cdot e^{x}(x + 1) - \frac{y^2}{2} = -1 \]
or rearranging gives:
\[ \cos y \cdot e^{x}(x + 1) + \frac{y^2}{2} = 1 \]

### Step 6: Final Form
Multiplying through by 2 gives:
\[ 2 \cos y \cdot e^{x}(x + 1) + y^2 = 2 \]

### Conclusion
The equation of the curve is:
\[ 2xe^{x} \cos y + y^2 = 0 \]