You have a differential equation involving two square root terms. Here's how we can interpret the given expression:

$\sqrt{1 - y^2} \, dx - \sqrt{2 - x^2} \, dy = 0, \quad y(0) = \sqrt{\frac{3}{2}}$

Step-by-step breakdown:

1. Separate variables: First, we'll try to separate the terms involving $x$ and $y$. $\sqrt{1 - y^2} \, dx = \sqrt{2 - x^2} \, dy$ Divide both sides to get: $\frac{dx}{\sqrt{2 - x^2}} = \frac{dy}{\sqrt{1 - y^2}}$

2. Integrate both sides: We'll integrate each side separately.

   • The integral of the left-hand side, $\int \frac{dx}{\sqrt{2 - x^2}}$, is of the form $\sin^{-1}(x/\sqrt{2})$.
   • The integral of the right-hand side, $\int \frac{dy}{\sqrt{1 - y^2}}$, is $\sin^{-1}(y)$.

   So, after integrating: $\sin^{-1}\left(\frac{x}{\sqrt{2}}\right) = \sin^{-1}(y) + C$

3. Solve for the constant $C$: Use the initial condition $y(0) = \sqrt{\frac{3}{2}}$. Plugging $x = 0$ and $y = \sqrt{\frac{3}{2}}$ into the equation: $\sin^{-1}\left(0\right) = \sin^{-1}\left(\sqrt{\frac{3}{2}}\right) + C$ Since $\sin^{-1}(0) = 0$, we have: $0 = \sin^{-1}\left(\sqrt{\frac{3}{2}}\right) + C$ Therefore, $C = -\sin^{-1}\left(\sqrt{\frac{3}{2}}\right)$.

4. Final solution: Substitute $C$ back into the equation: $\sin^{-1}\left(\frac{x}{\sqrt{2}}\right) = \sin^{-1}(y) - \sin^{-1}\left(\sqrt{\frac{3}{2}}\right)$

This is the implicit form of the solution.

Would you like to see the solution steps in more detail, or have any specific questions?

Related questions:

1. How do you solve integrals involving square roots of quadratic terms?
2. What is the geometric interpretation of inverse sine functions?
3. How does the initial condition affect the solution of differential equations?
4. How do you compute inverse trigonometric functions for specific values?
5. Can this differential equation be solved explicitly for $x$ or $y$?

Tip:

When solving differential equations, using initial conditions early can help simplify the process significantly.