Question

2x(1-y^2)dx-2y(1-x^2)dy=0​

Answer 1

Ответ:

x²-1=C(y²-1)

Пошаговое объяснение:

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Объединяем решения:

y=±1 , x=±1 , x²-1=c1(y²-1),c1≠0

и получаем общее решение, где С - произвольное:

x²-1=C(y²-1)

Answer 2

The given equation is a first-order ordinary differential equation (ODE). To solve it, we can use the method of separation of variables. Here's the step-by-step solution:

1. Rearrange the equation to separate the variables dx and dy: 2x(1 - y^2) dx = 2y(1 - x^2) dy

2. Divide both sides of the equation by 2x(1 - y^2) and multiply both sides by dy: dx/(x(1 - y^2)) = dy/(y(1 - x^2))

3. Integrate both sides with respect to their respective variables: ∫ dx/(x(1 - y^2)) = ∫ dy/(y(1 - x^2))

4. Evaluate the integrals: ∫ dx/(x(1 - y^2)) = ∫ dy/(y(1 - x^2)) ln|x| + ln|1 - y^2| = ln|y| + ln|1 - x^2| + C

   Here, C is the constant of integration.

5. Combine the logarithms using properties of logarithms: ln|x(1 - y^2)| = ln|y(1 - x^2)| + C

6. Exponentiate both sides to eliminate the natural logarithms: x(1 - y^2) = y(1 - x^2) * e^C

7. Rewrite e^C as a new constant, let's say K: x(1 - y^2) = K * y(1 - x^2)

   Here, K = e^C.

8. Simplify the equation: x - xy^2 = Ky - Kx^2

9. Rearrange the terms: Kx^2 - x + Ky - xy^2 = 0

10. This equation represents a quadratic equation in terms of x. We can solve it by applying the quadratic formula: x = (1 ± √(1 - 4K(Ky - xy^2))) / (2K)

Note: The discriminant (1 - 4K(Ky - xy^2)) must be non-negative for real solutions.

So, the solution to the given differential equation is x = (1 ± √(1 - 4K(Ky - xy^2))) / (2K), where K is a constant determined by the initial conditions or boundary conditions of the problem.