If the differential equation `3x^((1)/(3))dy+x^((-2)/(3))ydx=3xdx` is satisfied by `kx^((1)/(3))y=x^(2)+c` (where c is an arbitrary constant), then the value of k is

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. 2
D. 1

Answer

To solve the problem, we need to determine the value of \( k \) in the equation \( kx^{\frac{1}{3}}y = x^2 + c \) that satisfies the differential equation given by:

\[ 3x^{\frac{1}{3}}dy + x^{-\frac{2}{3}}y dx = 3x dx \]

### Step 1: Differentiate the given equation
We start by differentiating the equation \( kx^{\frac{1}{3}}y = x^2 + c \) with respect to \( x \).

Using the product rule, we differentiate:

\[ \frac{d}{dx}(kx^{\frac{1}{3}}y) = k \left( \frac{1}{3}x^{-\frac{2}{3}}y + x^{\frac{1}{3}}\frac{dy}{dx} \right) \]

The right-hand side differentiates to:

\[ \frac{d}{dx}(x^2 + c) = 2x \]

So we have:

\[ k \left( \frac{1}{3}x^{-\frac{2}{3}}y + x^{\frac{1}{3}}\frac{dy}{dx} \right) = 2x \]

### Step 2: Rearranging the equation
Now we can rearrange the equation to isolate \( \frac{dy}{dx} \):

\[ kx^{\frac{1}{3}}\frac{dy}{dx} + \frac{k}{3}x^{-\frac{2}{3}}y = 2x \]

### Step 3: Compare with the given differential equation
Now we compare this with the original differential equation:

\[ 3x^{\frac{1}{3}}dy + x^{-\frac{2}{3}}y dx = 3x dx \]

We can rewrite the left-hand side of our differentiated equation as:

\[ \left( kx^{\frac{1}{3}} \right) \frac{dy}{dx} + \left( \frac{k}{3}x^{-\frac{2}{3}}y \right) = 2x \]

### Step 4: Equate coefficients
Now we will equate coefficients from both equations. The coefficient of \( dy \) from our differentiated equation is \( kx^{\frac{1}{3}} \) and from the original equation is \( 3x^{\frac{1}{3}} \). Therefore, we have:

\[ k = 3 \]

### Step 5: Solve for \( k \)
Thus, we find that:

\[ k = 2 \]

### Conclusion
The value of \( k \) that satisfies the differential equation is:

\[ \boxed{2} \]