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 $\frac{dy}{dx} = \sqrt{1 - y^2} = \cos x$.

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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.