The differential equation for `x^(3)+y^(3)=4ax` is

Options

A. `3xy^(2)(dy)/(dx)+2x^(3)+y^(3)=0`
B. `3xy^(2)(dy)/(dx)+2x^(3)-y^(3)=0`
C. `3xy^(2)(dy)/(dx)-2x^(3)+y^(3)=0`
D. `3xy^(2)(dy)/(dx)-2x^(3)-y^(3)=0`

Answer

To find the differential equation for the curve defined by the equation \( x^3 + y^3 = 4ax \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x^3 + y^3 = 4ax \]
We can rearrange this to express \( y^3 \) in terms of \( x \) and \( a \):
\[ y^3 = 4ax - x^3 \]

### Step 2: Differentiate both sides
Now, we differentiate both sides of the equation with respect to \( x \):
\[ \frac{d}{dx}(y^3) = \frac{d}{dx}(4ax - x^3) \]
Using the chain rule on the left side and the product rule on the right side, we get:
\[ 3y^2 \frac{dy}{dx} = 4a - 3x^2 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Now, we will solve for \(\frac{dy}{dx}\):
\[ 3y^2 \frac{dy}{dx} = 4a - 3x^2 \]
Dividing both sides by \(3y^2\):
\[ \frac{dy}{dx} = \frac{4a - 3x^2}{3y^2} \]

### Step 4: Eliminate the parameter \( a \)
Next, we need to eliminate the parameter \( a \) from our expression. From the original equation \( x^3 + y^3 = 4ax \), we can express \( a \) as:
\[ a = \frac{x^3 + y^3}{4x} \]
Now, substituting this expression for \( a \) back into our derivative:
\[ \frac{dy}{dx} = \frac{4\left(\frac{x^3 + y^3}{4x}\right) - 3x^2}{3y^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{x^3 + y^3 - 3x^3}{3xy^2} = \frac{y^3 - 2x^3}{3xy^2} \]

### Step 5: Form the differential equation
Now we can rearrange this to form the differential equation:
\[ 3xy^2 \frac{dy}{dx} + 2x^3 - y^3 = 0 \]

### Final Result
Thus, the differential equation for the given curve \( x^3 + y^3 = 4ax \) is:
\[ 3xy^2 \frac{dy}{dx} + 2x^3 - y^3 = 0 \]