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

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the given differential equation \( x^2 dy + 2xy dx = \sin x dx \) and find the value of \( k \) in the solution form \( x^k y + \cos x = C \), we can follow these steps:

### Step 1: Rearranging the Differential Equation
We start with the equation:
\[ x^2 dy + 2xy dx = \sin x dx \]
We can rearrange this to:
\[ x^2 dy = \sin x dx - 2xy dx \]
This simplifies to:
\[ x^2 dy = (\sin x - 2xy) dx \]

### Step 2: Identifying the Exact Differential
Notice that the left-hand side \( x^2 dy + 2xy dx \) can be recognized as the total differential of the function \( x^2 y \). We can express this as:
\[ d(x^2 y) = \sin x dx \]

### Step 3: Integrating Both Sides
Now, we integrate both sides:
\[ \int d(x^2 y) = \int \sin x \, dx \]
The left-hand side simplifies to:
\[ x^2 y \]
The right-hand side integrates to:
\[ -\cos x + C \]
Thus, we have:
\[ x^2 y = -\cos x + C \]

### Step 4: Rearranging the Result
Rearranging gives us:
\[ x^2 y + \cos x = C \]

### Step 5: Comparing with the Given Solution Form
The problem states that the solution can be expressed as:
\[ x^k y + \cos x = C \]
From our derived solution, we see that:
\[ x^2 y + \cos x = C \]
This implies that \( k = 2 \).

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{2} \]