Question

(2x+3x^2 y)dx+(x^3-3y^2)dy=0​

Answer 1

ответ:

2dx^2+4dx^3*y*3dy^3

и кстати, это разве относится к географии?)

Answer 2

To solve the given first-order ordinary differential equation (ODE):

(2x + 3x^2y)dx + (x^3 - 3y^2)dy = 0

We can use the method of exact differential equations. For an exact differential equation, there exists a function called the potential function (or simply the potential) whose partial derivatives with respect to x and y will match the coefficients of dx and dy, respectively.

Step 1: Checking for Exactness To determine if the equation is exact, check if the mixed partial derivatives are equal:

∂/∂y(2x + 3x^2y) = 3x^2 ∂/∂x(x^3 - 3y^2) = 3x^2

Since the mixed partial derivatives are equal (3x^2 = 3x^2), the equation is exact.

Step 2: Finding the Potential Function Now, we need to find the potential function F(x, y) such that:

∂F/∂x = 2x + 3x^2y ∂F/∂y = x^3 - 3y^2

Integrate the first equation with respect to x to find F(x, y):

∫(2x + 3x^2y)dx = x^2 + x^3y + g(y)

Here, g(y) is an arbitrary function of y. Now, differentiate the result with respect to y to find the partial derivative with respect to y:

∂/∂y(x^2 + x^3y + g(y)) = x^3 + g'(y)

Comparing this with ∂F/∂y = x^3 - 3y^2, we get:

x^3 + g'(y) = x^3 - 3y^2

So, g'(y) = -3y^2

Integrate g(y) to find g'(y):

∫(-3y^2)dy = -y^3 + C

where C is an integration constant. Now, we can write the potential function F(x, y) as:

F(x, y) = x^2 + x^3y - y^3 + C

Step 3: Solution The solution to the given differential equation is the equation of the potential function F(x, y) obtained in Step 2:

x^2 + x^3y - y^3 = C

where C is an arbitrary constant of integration. This represents the implicit solution to the given ODE.