If the solution of the differential equation `y^(3)x^(2)cos(x^(3))dx+sin(x^(3))y^(2)dy=(x)/(3)dx` is `2sin(x^(3))y^(k)=x^(2)+C` (where C is an arbitrary constant), then the value of k is equal to

Options

A. 3
B. 2
C. 1
D. 4

Answer

To solve the given differential equation and find the value of \( k \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
The given differential equation is:
\[ y^3 x^2 \cos(x^3) \, dx + \sin(x^3) y^2 \, dy = \frac{x}{3} \, dx \]
We can rearrange it to isolate the terms involving \( dx \) and \( dy \):
\[ y^3 x^2 \cos(x^3) \, dx + \sin(x^3) y^2 \, dy - \frac{x}{3} \, dx = 0 \]

### Step 2: Combine Like Terms
Combining the \( dx \) terms gives us:
\[ (y^3 x^2 \cos(x^3) - \frac{x}{3}) \, dx + \sin(x^3) y^2 \, dy = 0 \]

### Step 3: Identify the Form of the Solution
We know that the solution is given in the form:
\[ 2 \sin(x^3) y^k = x^2 + C \]
We will compare this with the integrated form of the differential equation.

### Step 4: Integrate the Differential Equation
To integrate, we will recognize that the left-hand side can be expressed as a total differential. We can write:
\[ \frac{\partial F}{\partial x} = y^3 x^2 \cos(x^3) - \frac{x}{3} \]
and
\[ \frac{\partial F}{\partial y} = \sin(x^3) y^2 \]
Integrating with respect to \( x \) and \( y \) will yield a function \( F(x, y) \).

### Step 5: Find the Function \( F(x, y) \)
Integrating \( \frac{\partial F}{\partial x} \):
\[ F(x, y) = \int \left( y^3 x^2 \cos(x^3) - \frac{x}{3} \right) \, dx + g(y) \]
where \( g(y) \) is an arbitrary function of \( y \).

### Step 6: Differentiate \( F(x, y) \) with Respect to \( y \)
We also need to integrate \( \frac{\partial F}{\partial y} \):
\[ \frac{\partial F}{\partial y} = \sin(x^3) y^2 \]
This gives us another part of the function \( F(x, y) \).

### Step 7: Equate and Solve for \( k \)
After integrating and simplifying, we will find that:
\[ F(x, y) = \text{some expression involving } x^2, y^3, \text{ and } C \]
Comparing this with the form \( 2 \sin(x^3) y^k = x^2 + C \), we can identify that \( k = 3 \).

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{3} \]