The solution of the differential equation `2sqrtxe^(sqrtx)dy+e^(sqrtx)ydx=sqrtx sin xdx` is (where, c is arbitrary constant)

Options

A. `2ye^(sqrtx)+sinx=c`
B. `ysinx=e^(sqrtx)+c`
C. `ye^(sqrtx)+sinx=c`
D. `2ye^(sqrtx)+cosx=c`

Answer

To solve the given differential equation \( 2\sqrt{x} e^{\sqrt{x}} dy + e^{\sqrt{x}} y dx = \sqrt{x} \sin x dx \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by dividing the entire equation by \( 2\sqrt{x} \):
\[ e^{\sqrt{x}} dy + \frac{e^{\sqrt{x}}}{2\sqrt{x}} y dx = \frac{\sqrt{x} \sin x}{2\sqrt{x}} dx \]
This simplifies to:
\[ e^{\sqrt{x}} dy + \frac{e^{\sqrt{x}}}{2\sqrt{x}} y dx = \frac{1}{2} \sin x dx \]

### Step 2: Identifying the Form
Notice that the left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(y e^{\sqrt{x}}) = e^{\sqrt{x}} dy + y \frac{d}{dx}(e^{\sqrt{x}}) \]
Since \( \frac{d}{dx}(e^{\sqrt{x}}) = \frac{1}{2\sqrt{x}} e^{\sqrt{x}} \), we can rewrite the left-hand side:
\[ \frac{d}{dx}(y e^{\sqrt{x}}) = \frac{1}{2} \sin x \]

### Step 3: Integrating Both Sides
Now we integrate both sides with respect to \( x \):
\[ \int \frac{d}{dx}(y e^{\sqrt{x}}) dx = \int \frac{1}{2} \sin x \, dx \]
This gives us:
\[ y e^{\sqrt{x}} = -\frac{1}{2} \cos x + C \]
where \( C \) is the constant of integration.

### Step 4: Solving for \( y \)
Now, we can solve for \( y \):
\[ y = \frac{-\frac{1}{2} \cos x + C}{e^{\sqrt{x}}} \]
This can be rewritten as:
\[ y = \frac{-\cos x}{2 e^{\sqrt{x}}} + \frac{C}{e^{\sqrt{x}}} \]

### Final Solution
Thus, the solution of the differential equation is:
\[ y = \frac{-\cos x}{2 e^{\sqrt{x}}} + C e^{-\sqrt{x}} \]