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

### Step 1: Rewrite the Equation
We start by rewriting the equation in a more manageable form. We divide the entire equation by \( 2\sqrt{x} \):

\[ e^{\sqrt{x}} dy + \frac{e^{\sqrt{x}} y}{2\sqrt{x}} dx = \frac{\sqrt{x} \sin x}{2\sqrt{x}} dx \]

This simplifies to:

\[ e^{\sqrt{x}} dy + \frac{e^{\sqrt{x}} y}{2\sqrt{x}} dx = \frac{1}{2} \sin x dx \]

### Step 2: Identify the Exact Form
We can express the equation in the form \( M dy + N dx = 0 \), where:

- \( M = e^{\sqrt{x}} \)
- \( N = \frac{e^{\sqrt{x}} y}{2\sqrt{x}} - \frac{1}{2} \sin x \)

To check if the equation is exact, we need to verify if \( \frac{\partial M}{\partial x} = \frac{\partial N}{\partial y} \).

### Step 3: Calculate Partial Derivatives
Calculating the partial derivatives:

- \( \frac{\partial M}{\partial x} = \frac{1}{2} e^{\sqrt{x}} \cdot \frac{1}{\sqrt{x}} \)
- \( \frac{\partial N}{\partial y} = \frac{e^{\sqrt{x}}}{2\sqrt{x}} \)

Since both derivatives are equal, the equation is exact.

### Step 4: Find the Potential Function
To find the potential function \( \Psi(x, y) \), we integrate \( M \) with respect to \( y \):

\[ \Psi(x, y) = \int e^{\sqrt{x}} dy = e^{\sqrt{x}} y + g(x) \]

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

Next, we differentiate \( \Psi \) with respect to \( x \):

\[ \frac{\partial \Psi}{\partial x} = \frac{1}{2\sqrt{x}} e^{\sqrt{x}} y + g'(x) \]

Setting this equal to \( N \):

\[ \frac{1}{2\sqrt{x}} e^{\sqrt{x}} y + g'(x) = \frac{e^{\sqrt{x}} y}{2\sqrt{x}} - \frac{1}{2} \sin x \]

This implies:

\[ g'(x) = -\frac{1}{2} \sin x \]

### Step 5: Integrate to Find \( g(x) \)
Integrating \( g'(x) \):

\[ g(x) = -\frac{1}{2} \int \sin x \, dx = \frac{1}{2} \cos x + C \]

### Step 6: Write the Complete Solution
Thus, the potential function is:

\[ \Psi(x, y) = e^{\sqrt{x}} y + \frac{1}{2} \cos x + C = 0 \]

### Final Solution
Rearranging gives us the final solution:

\[ 2e^{\sqrt{x}} y + \cos x = C \]