Question

Sqrt(1 - y ^ 2) * dx + sqrt(1 - x ^ 2) * dy = 0

Answer 1

Решение .

Дифференциальное уравнение 1 порядка с разделяющимися переменными .

$\bf \displaystyle \sqrt{1-y^2}\, dx+\sqrt{1-x^2}\, dy=0\\\\\sqrt{1-y^2}\, dx=-\sqrt{1-x^2}\, dy\\\\\int \frac{dy}{\sqrt{1-y^2}}=-\int \frac{dx}{\sqrt{1-x^2}}\\\\\\arcsiny=-arcsinx+arcsinC\\\\y=sin(arcsinC-arcsinx)\\\\y=C\cdot \sqrt{1-x^2}-x\cdot \sqrt{1-C^2}$  

Answer 2

The equation you provided is a differential equation known as a "differential form." Specifically, it represents a differential form on the unit circle in the xy-plane.

To solve this equation, we can start by rearranging the terms and isolating the variables:

sqrt(1 - y^2) * dx = -sqrt(1 - x^2) * dy

Now, let's integrate both sides of the equation. Since the equation represents a differential form on the unit circle, we can integrate it along a path on the unit circle.

∫ sqrt(1 - y^2) * dx = ∫ -sqrt(1 - x^2) * dy

The integral on the left-hand side can be evaluated by treating y as a constant, and the integral on the right-hand side can be evaluated by treating x as a constant. The integrals will give us the antiderivatives of the respective functions. Let's perform the integrations:

∫ sqrt(1 - y^2) * dx = -∫ sqrt(1 - x^2) * dy

Let F(x) be the antiderivative of sqrt(1 - y^2) with respect to x, and G(y) be the antiderivative of -sqrt(1 - x^2) with respect to y. The equation becomes:

F(x) = -G(y) + C

where C is the constant of integration.

Now, we can solve this equation for x in terms of y or y in terms of x, depending on which variable we want to express in terms of the other.

Let's solve for x in terms of y:

F(x) = -G(y) + C

x = F^(-1)(-G(y) + C)

Here, F^(-1) represents the inverse function of F.

Similarly, we can solve for y in terms of x:

F(x) = -G(y) + C

y = G^(-1)(-F(x) + C)

Again, G^(-1) represents the inverse function of G.

Please note that without specific functions for sqrt(1 - y^2) and -sqrt(1 - x^2), we cannot determine the explicit solutions. The solutions would depend on the specific form of these functions.