Question

\(x \sqrt{1-y^{2}} d x+y \sqrt{1-x^{2}} d y=0, y(0)=1\).

Short answer

Answer: The implicit solution for the given exact differential equation is \(x^2\ \frac{\sqrt{1-y^2}}{2} + y^2\ \frac{\sqrt{1-x^2}}{2}- \frac{1}{2} = 0\).

Step-by-step solution

Check if the given differential equation is exact

To check if the given exact differential equation is exact, first calculate the partial derivatives: \(\frac{\partial}{\partial y}( x \sqrt{1-y^{2}} )= -x y \ \frac{1}{\sqrt{1-y^{2}}}\) \(\frac{\partial}{\partial x}( y \sqrt{1-x^{2}} )= -x y \ \frac{1}{\sqrt{1-x^{2}}}\) Since \(\frac{\partial}{\partial y}( x \sqrt{1-y^{2}} ) = \frac{\partial}{\partial x}( y \sqrt{1-x^{2}} )\), the differential equation is exact.

Identify the potential function ψ(x, y)

To find the potential function ψ(x, y) that satisfies the exact differential equation, we need to integrate the given expressions with respect to their respective variables: Integrate with respect to x: \(\int x \sqrt{1-y^{2}}dx = x^2\ \frac{\sqrt{1-y^2}}{2} + C_1(y)\) Integrate with respect to y: \(\int y \sqrt{1-x^{2}}dy = y^2\ \frac{\sqrt{1-x^2}}{2} + C_2(x)\)

Determine the constants of integration

Now, we have \(\psi(x, y) = x^2\ \frac{\sqrt{1-y^2}}{2} + y^2\ \frac{\sqrt{1-x^2}}{2} + C(x, y)\) The potential function, however, has no constants of integration, so we simply have: \(\psi(x, y) = x^2\ \frac{\sqrt{1-y^2}}{2} + y^2\ \frac{\sqrt{1-x^2}}{2}\)

Apply the initial condition

Now we apply the given initial condition \(y(0) = 1\): \(\psi(0, 1) = 0^2\ \frac{\sqrt{1-1^2}}{2} + 1^2\ \frac{\sqrt{1-0^2}}{2}\) \(\psi(0, 1) = 0 + \frac{1}{2}\) Which gives us: \(\psi(x, y) - \frac{1}{2} = 0\).

Write down the final implicit solution

Finally, we can rewrite the above expression for the implicit solution of the exact differential equation: \(x^2\ \frac{\sqrt{1-y^2}}{2} + y^2\ \frac{\sqrt{1-x^2}}{2}- \frac{1}{2} = 0\) This is the implicit solution to the given exact differential equation, considering the initial condition \(y(0) = 1\).