Question

X(1-y^2)dx+y(1-x^2)dy=0

Answer 1

ДУ с разделяющимися переменными:

$\displaystyle x(1-y^2)dx+y(1-x^2)dy=0\\x(1-y^2)dx=-y(1-x^2)dy|*\frac{1}{(1-x^2)(1-y^2)}\\\frac{xdx}{1-x^2}=-\frac{ydy}{1-y^2}\\-\int\frac{d(1-x^2)}{1-x^2}=\int\frac{(1-y^2)dy}{1-y^2}\\-ln|1-x^2|=ln|1-y^2|+C\\\frac{1}{1-x^2}=C(1-y^2)\\(1-x^2)(1-y^2)=C;y=^+_-1;x=^+_-1$

Answer 2

The given equation is:

X(1 - y^2)dx + y(1 - x^2)dy = 0

To solve this first-order ordinary differential equation, we can check if it is exact or not. If it is exact, we can find a potential function whose partial derivatives with respect to x and y match the coefficients of dx and dy in the equation.

To determine if the equation is exact, we compute the partial derivatives of the coefficients with respect to x and y:

∂/∂y [X(1 - y^2)] = -2Xy ∂/∂x [y(1 - x^2)] = -2xy

Since these two partial derivatives are not equal, the equation is not exact.

To solve it, we can try to find an integrating factor, denoted by μ(x, y), which multiplies both sides of the equation to make it exact. The integrating factor is given by:

μ(x, y) = e^(∫((∂/∂y)[y(1 - x^2)] - (∂/∂x)[X(1 - y^2)])dx)

Calculating the integrating factor:

∂/∂y [y(1 - x^2)] - ∂/∂x [X(1 - y^2)] = (1 - x^2) - 2xy = 1 - x^2 - 2xy

∫(1 - x^2 - 2xy)dx = ∫(1 - x^2 - 2xy)dx = x - (1/3)x^3 - x^2y

μ(x, y) = e^(x - (1/3)x^3 - x^2y)

Multiplying the original equation by the integrating factor:

e^(x - (1/3)x^3 - x^2y) * (X(1 - y^2)dx + y(1 - x^2)dy) = 0

Now, we need to check if the new equation is exact. To do this, we calculate the partial derivatives of the new coefficients with respect to x and y:

∂/∂y [e^(x - (1/3)x^3 - x^2y) * y(1 - x^2)] = e^(x - (1/3)x^3 - x^2y)(1 - x^2 - 2xy) - 2x^2ye^(x - (1/3)x^3 - x^2y) ∂/∂x [e^(x - (1/3)x^3 - x^2y) * X(1 - y^2)] = e^(x - (1/3)x^3 - x^2y)(1 - x^2 - 2xy) - e^(x - (1/3)x^3 - x^2y)(1 - y^2)

The equation is exact if these two partial derivatives are equal. Let's check:

e^(x - (1/3)x^3 - x^2y)(1 - x^2 - 2xy) - 2x^2ye^(x - (1/3)x^3 - x^2y) = e^(x - (1/3)x^3 - x^2y)(1 - x^2 - 2xy) - e^(x - (1/3)x^3 - x^2y)(1 - y^2)

The partial derivatives are indeed equal, which means the equation is now exact. We can find a potential function, denoted by φ(x, y), such that:

∂φ/∂x = e^(x - (1/3)x^3 - x^2y)(1 - y^2) and ∂φ/∂y = e^(x - (1/3)x^3 - x^2y)(1 - x^2 - 2xy)

To find φ(x, y), we integrate the first expression with respect to x:

φ(x, y) = ∫e^(x - (1/3)x^3 - x^2y)(1 - y^2)dx

To solve this integral, we need to perform further calculations. However, due to the complexity of the equation, it may not have a simple closed-form solution.

Therefore, the solution to the given differential equation is given implicitly by the potential function φ(x, y) above.