Question

2xcos^2ydx+(2y-x^2sin2y)dy=0

Answer 1

d(y2+x2cos2y)y2+x2cos2y=C

(все двойки пиши с верху:)

Answer 2

To solve the given first-order differential equation:

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

We can start by checking if it is an exact differential equation. An exact equation is one in which the following condition is satisfied:

∂M/∂y = ∂N/∂x

Where M is the coefficient of dx and N is the coefficient of dy in the equation.

In our equation:

M = 2xcos^2(y) N = 2y - x^2sin(2y)

Now, let's calculate the partial derivatives:

∂M/∂y = -2xsin(2y) ∂N/∂x = -2xsin(2y)

Both ∂M/∂y and ∂N/∂x are equal, which means this is an exact differential equation. To solve it, we need to find a potential function Φ(x, y) such that:

∂Φ/∂x = M ∂Φ/∂y = N

Let's integrate M with respect to x and N with respect to y separately to find Φ:

∫2xcos^2(y)dx = x^2cos^2(y) + g(y), where g(y) is a function of y only.

∫(2y - x^2*sin(2y))dy = y^2 - (1/4)x^2cos(2y) + h(x), where h(x) is a function of x only.

Now, we equate the two expressions for Φ:

x^2*cos^2(y) + g(y) = y^2 - (1/4)x^2cos(2y) + h(x)

To find g(y) and h(x), we need to equate the parts that depend only on y and x, respectively:

g(y) = y^2 h(x) = - (1/4)x^2cos(2y)

Now, let's find g(y) and h(x):

g(y) = y^2 ∂g/∂y = 2y

h(x) = - (1/4)x^2cos(2y) ∂h/∂x = - (1/4)2x*cos(2y) = - (1/2)xcos(2y)

Now, equate these partial derivatives to the respective partial derivatives of Φ:

∂Φ/∂x = - (1/2)xcos(2y) ∂Φ/∂y = 2y

To make these equal to the partial derivatives of Φ, we need to choose g(y) = y^2 and h(x) = - (1/2)xcos(2y).

So, our potential function Φ(x, y) is given by:

Φ(x, y) = x^2*cos^2(y) + y^2 - (1/2)x^2cos(2y)

Now, we have found the potential function Φ(x, y), and the solution to the original differential equation is:

Φ(x, y) = C

Where C is the constant of integration.