Solution of the differential equation
`x((dy)/(dx))^(2)+2sqrt(xy)(dy)/(dx)+y=0`,is

Options

A. `x+y=a`
B. `sqrtx-sqrty=a`
C. `x^(2)+y^(2)=a^(2)`
D. `sqrtx+sqrty=sqrta`

Explanation

To solve the differential equation

\[ x\left(\frac{dy}{dx}\right)^{2} + 2\sqrt{xy}\frac{dy}{dx} + y = 0, \]

we can start by rewriting the equation in a more manageable form.

### Step 1: Rewrite the equation
We can express the equation as:

\[ x\left(\frac{dy}{dx}\right)^{2} + 2\sqrt{xy}\frac{dy}{dx} + y = 0. \]

This resembles the form of a quadratic equation in terms of \(\frac{dy}{dx}\). Let \(z = \frac{dy}{dx}\). Then we have:

\[ xz^{2} + 2\sqrt{xy}z + y = 0. \]

### Step 2: Apply the quadratic formula
Using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = x\), \(b = 2\sqrt{xy}\), and \(c = y\):

\[ z = \frac{-2\sqrt{xy} \pm \sqrt{(2\sqrt{xy})^2 - 4xy}}{2x}. \]

### Step 3: Simplify the expression
Calculating the discriminant:

\[ (2\sqrt{xy})^2 - 4xy = 4xy - 4xy = 0. \]

Since the discriminant is zero, we have one repeated root:

\[ z = \frac{-2\sqrt{xy}}{2x} = -\frac{\sqrt{y}}{\sqrt{x}}. \]

### Step 4: Substitute back for \(\frac{dy}{dx}\)
Thus, we have:

\[ \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}. \]

### Step 5: Separate variables
Rearranging gives:

\[ \sqrt{y} \, dy = -\sqrt{x} \, dx. \]

### Step 6: Integrate both sides
Integrating both sides:

\[ \int \sqrt{y} \, dy = -\int \sqrt{x} \, dx. \]

The left side integrates to:

\[ \frac{2}{3}y^{3/2} + C_1, \]

and the right side integrates to:

\[ -\frac{2}{3}x^{3/2} + C_2. \]

### Step 7: Combine the constants
Setting \(C = C_2 - C_1\), we have:

\[ \frac{2}{3}y^{3/2} + \frac{2}{3}x^{3/2} = C. \]

### Step 8: Final form of the solution
Multiplying through by \(\frac{3}{2}\) gives:

\[ y^{3/2} + x^{3/2} = C'. \]

This is the general solution of the differential equation.

### Summary of the solution
The solution of the differential equation

\[ x\left(\frac{dy}{dx}\right)^{2} + 2\sqrt{xy}\frac{dy}{dx} + y = 0 \]

is

\[ y^{3/2} + x^{3/2} = C, \]

where \(C\) is a constant.