The given differential equation is:

$\sqrt{1-x^2} \, dy - y \, dx = 0$

This can be rewritten as:

$\sqrt{1 - x^2} \, dy = y \, dx$

To solve this differential equation, we can first rewrite it in a separable form:

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

Now, integrate both sides:

$\int \frac{1}{y} \, dy = \int \frac{1}{\sqrt{1 - x^2}} \, dx$

Left-hand side:

$\int \frac{1}{y} \, dy = \ln |y|$

Right-hand side:

The integral of $\frac{1}{\sqrt{1 - x^2}}$ is the inverse sine function, so: $\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin(x)$

Now, combining both integrals:

$\ln |y| = \arcsin(x) + C$

Exponentiate both sides to solve for $y$:

$|y| = e^{\arcsin(x) + C} = e^C e^{\arcsin(x)}$

Let $A = e^C$, a constant. So, we have:

$y = A e^{\arcsin(x)}$

Thus, the general solution to the differential equation is:

$y = A e^{\arcsin(x)}$

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Would you like more details on any step or have any further questions?

Here are five related questions for further exploration:

1. How do we find the particular solution if an initial condition is provided?
2. Can we use a different method to solve this differential equation, such as an exact method?
3. What is the domain of the solution $y = A e^{\arcsin(x)}$?
4. How does the inverse sine function behave near its critical points?
5. How would we solve if the equation were slightly different, such as $\sqrt{1 - x^2} dy + y dx = 0$?

Tip: Remember, inverse trigonometric functions often arise in differential equations involving square roots of quadratic expressions.