The solution of the differential equation `sinye^(x)dx-e^(x)cos ydy=sin^(2)ydx` is (where, c is an arbitrary constant)

Options

A. `e^(x)siny=x+c`
B. `e^(x)=(x+c)siny`
C. `e^(x).x=siny+c`
D. `e^(x).siny=x^(2)+c`

Answer

To solve the differential equation \( \sin y e^x \, dx - e^x \cos y \, dy = \sin^2 y \, dx \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given differential equation:
\[ \sin y e^x \, dx - e^x \cos y \, dy - \sin^2 y \, dx = 0 \]
This can be rewritten as:
\[ (\sin y e^x - \sin^2 y) \, dx - e^x \cos y \, dy = 0 \]

### Step 2: Factor Out Common Terms
Next, we can factor out \( e^x \) from the first term:
\[ e^x (\sin y - \frac{\sin^2 y}{e^x}) \, dx - e^x \cos y \, dy = 0 \]
This simplifies to:
\[ (\sin y - \frac{\sin^2 y}{e^x}) \, dx - \cos y \, dy = 0 \]

### Step 3: Dividing by \( e^x \)
Now, we can divide the entire equation by \( e^x \):
\[ \left(\frac{\sin y}{e^x} - \frac{\sin^2 y}{e^{2x}}\right) \, dx - \frac{\cos y}{e^x} \, dy = 0 \]

### Step 4: Identifying Variables
Let \( u = \sin y \) and \( v = e^x \). Then, we can express the equation in terms of \( u \) and \( v \):
\[ \frac{u}{v} \, dx - \frac{\cos y}{v} \, dy = 0 \]

### Step 5: Applying the Product Rule
This can be rewritten using the product rule:
\[ \frac{d}{dx} \left( \frac{v}{u} \right) = \frac{dx}{dy} \]

### Step 6: Integrating Both Sides
Integrating both sides gives:
\[ \frac{v}{u} = x + C \]
Substituting back \( v = e^x \) and \( u = \sin y \):
\[ \frac{e^x}{\sin y} = x + C \]

### Step 7: Final Rearrangement
Rearranging gives us the solution:
\[ e^x = (x + C) \sin y \]

### Final Solution
Thus, the solution of the differential equation is:
\[ e^x = (x + C) \sin y \]

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