The differental equation satisfied by the function ` y=sqrt(sin x +sqrt( sin x +sqrt(sin x + . . . + oo)))` is

Options

A. `y=(dy)/(dx)- sin x=0`
B. ` (2y -1) cos x+ (dy)/(dx)=0`
C. `(2y-1) cos x -(dy)/(dx)=0``(2y-1)(dy)/(dx)- cos x=0`
D. `(2y-1)(dy)/(dx) - cos x=0`

Answer

To find the differential equation satisfied by the function \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}} \), we can follow these steps:

### Step 1: Define the function
Let:
\[ y = \sqrt{\sin x + y} \]
This is because the expression inside the square root repeats infinitely.

### Step 2: Square both sides
Squaring both sides gives:
\[ y^2 = \sin x + y \]

### Step 3: Rearrange the equation
Rearranging the equation, we have:
\[ y^2 - y - \sin x = 0 \]

### Step 4: Differentiate with respect to \( x \)
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) - \frac{d}{dx}(y) - \frac{d}{dx}(\sin x) = 0 \]
Using the chain rule, we get:
\[ 2y \frac{dy}{dx} - \frac{dy}{dx} - \cos x = 0 \]

### Step 5: Factor out \( \frac{dy}{dx} \)
Factoring out \( \frac{dy}{dx} \) gives:
\[ \frac{dy}{dx}(2y - 1) - \cos x = 0 \]

### Step 6: Rearranging to find the differential equation
Rearranging this equation leads us to:
\[ \frac{dy}{dx}(2y - 1) = \cos x \]
Thus, we can express the differential equation as:
\[ \frac{dy}{dx} = \frac{\cos x}{2y - 1} \]

### Final Form of the Differential Equation
The final form of the differential equation satisfied by the function \( y \) is:
\[ \frac{dy}{dx} = \frac{\cos x}{2y - 1} \]

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