Given that y = sin x is an explicit solution of the first order differential equation
dy
dx =
√
1 − y
2
. Find the interval I where such equation is defined.

Given the differential equation:

\[
\frac{dy}{dx} = \sqrt{1 - y^2}
\]

we are to determine the interval \( I \) where \( y = \sin x \) is a solution. Let's first verify that \( y = \sin x \) is indeed a solution to this differential equation.

### Step-by-Step Solution:

1. **Verify \( y = \sin x \) is a Solution:**

   The derivative of \( y = \sin x \) with respect to \( x \) is:

   \[
   \frac{dy}{dx} = \cos x
   \]

   According to the given differential equation:

   \[
   \frac{dy}{dx} = \sqrt{1 - y^2}
   \]

   Substitute \( y = \sin x \) into the right-hand side:

   \[
   \sqrt{1 - (\sin x)^2} = \sqrt{1 - \sin^2 x}
   \]

   Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we have:

   \[
   \sqrt{1 - \sin^2 x} = \sqrt{\cos^2 x} = |\cos x|
   \]

   Therefore, the differential equation becomes:

   \[
   \frac{dy}{dx} = |\cos x|
   \]

   Since \( \cos x \) can be positive or negative depending on the interval of \( x \), the equality \( \frac{dy}{dx} = \cos x \) will hold true only if \( \cos x \) is non-negative (i.e., \( \cos x \geq 0 \)). This happens in the interval where \( x \) is within \( [2k\pi, (2k+1)\pi] \) for any integer \( k \).

### Determine the Interval \( I \):

For \( y = \sin x \) to be a valid solution, \( |\cos x| \) must equal \( \cos x \). This holds when \( \cos x \geq 0 \). Therefore, we find:

\[
\cos x \geq 0 \implies x \in [2k\pi, (2k+1)\pi] \text{ for integer } k
\]

### Conclusion:

The solution \( y = \sin x \) is defined in the interval:

\[
I = [2k\pi, (2k+1)\pi] \text{ for any integer } k
\]

This interval covers all ranges where \( \cos x \) is non-negative, ensuring that \( \frac{dy}{dx} = \sqrt{1 - y^2} = \cos x \).

---

Do you want details or have any questions?

Here are some related questions for further exploration:

1. How does the interval change if the initial condition of the differential equation is specified?
2. What is the behavior of the solution \( y = \sin x \) outside the interval \( [2k\pi, (2k+1)\pi] \)?
3. Can there be other explicit solutions to the differential equation \( \frac{dy}{dx} = \sqrt{1 - y^2} \)?
4. How would the solution change if the differential equation were \( \frac{dy}{dx} = -\sqrt{1 - y^2} \)?
5. What is the significance of the interval \( I \) in the context of real-world applications of this differential equation?

**Tip:** When verifying solutions to differential equations, always consider the domain and range of the functions involved to ensure the solution is valid over the specified interval.

Given that y = sin x is an explicit solution of the first order differential equation dy/dx = √(1 − y^2), find the interval I where such equation is defined.

Discover how to solve the differential equation dy/dx = √(1 − y^2) and find the interval where y = sin(x) is the solution. This problem involves trigonometry, implicit differentiation, and applying the Pythagorean identity. Suitable for advanced high school students.

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