The differential equation for the family of curves `y = c\ sinx` can be given by

Options

A. `((dy)/(dx))^(2) = y^(2)cot^(2)x`
B. `((dy)/(dx))^(2)-(sec x(dy)/(dx))^(2)+y^(2) = 0`
C. `((dy)/(dx))^(2)=tan^(2)x`
D. `(dy)/(dx)=y cot x`

Explanation

To find the differential equation for the family of curves given by \( y = c \sin x \), we can follow these steps:

### Step 1: Differentiate the given equation
We start with the equation:
\[ y = c \sin x \]
Differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = c \cos x \]

### Step 2: Express \( c \) in terms of \( y \) and \( x \)
From the first equation, we can express \( c \) as:
\[ c = \frac{y}{\sin x} \]

### Step 3: Substitute \( c \) into the derivative
Now we substitute \( c \) into the derivative we found in Step 1:
\[ \frac{dy}{dx} = \left(\frac{y}{\sin x}\right) \cos x \]
This simplifies to:
\[ \frac{dy}{dx} = y \frac{\cos x}{\sin x} = y \cot x \]

### Step 4: Rearranging the equation
We can rearrange this equation to isolate the derivative:
\[ \frac{dy}{dx} - y \cot x = 0 \]

### Step 5: Form the differential equation
The final form of the differential equation is:
\[ \frac{dy}{dx} = y \cot x \]

### Step 6: Verify the equation
To verify, we can check if the original family of curves satisfies this differential equation. Substituting \( y = c \sin x \) into the equation should hold true.

### Summary
The differential equation for the family of curves \( y = c \sin x \) is:
\[ \frac{dy}{dx} = y \cot x \]

---