Question

(2x sin y + 3x²) dx + (x ² cos y + 1/y) dy = 0​

Answer 1

Ответ:

$x^2 \sin y+x^3+\ln |y|=C$

Объяснение:

Дано уравнение вида:

Pdx+Qdy=0, где P=2x sin y + 3x², Q=x² cos y + 1/y

Найдем следующие частные производные:

$P'_y=(2x \sin y + 3x^2)'_y=2x \cos y \\ \\ Q'_x=(x^2 \cos y + \frac{1}{y} )'_x=2x\cos y$

Так как $P'_y=Q'_x$ то дано уравнение в полных дифференциалах.

Решение:

$Pdx+Qdy=0 \ \Rightarrow \ du=0 \ \Rightarrow \ u=C$

Задача сводится к тому, чтобы найти неизвестную функцию u(x,y).

$du=u'_xdx+u'_ydy$

С другой стороны $du=Pdx+Qdy$, тогда

$u'_x=P \ \Rightarrow \ u'_x=2x \sin y+3x^2 \ \Rightarrow \ u=\int (2x \sin y+3x^2)dx= \\ \\ x^2 \sin y+x^3+f(y)\\ \\ u'_y=(x^2 \sin y+x^3+f(y))'_y=x^2 \cos y +f'(y)$

С другой стороны:

$u'_y=Q \ \Rightarrow \ x^2 \cos y+f'(y)=x^2 \cos y+\frac{1}{y} \ \Rightarrow \ f'(y)=\frac{1}{y} \ \Rightarrow \ \\ \\ \Rightarrow \ f(y)=\int \frac{1}{y} \ dy=\ln |y|+C$

Таким образом:

$u=x^2 \sin y+x^3+f(y)=x^2 \sin y+x^3+\ln |y| +C$

Answer 2

The given equation is a first-order ordinary differential equation of the form:

(2xsin(y) + 3x^2)dx + (x^2cos(y) + 1/y)dy = 0

To solve this equation, we can use the method of exact differentials. For an equation to be exact, the partial derivatives with respect to x and y should be equal. Let's check that:

∂/∂x (2xsin(y) + 3x^2) = 2sin(y) + 6x ∂/∂y (x^2cos(y) + 1/y) = -x^2sin(y) - 1/y^2

The equation is not exact as the partial derivatives are not equal. However, we can try to find an integrating factor (μ) to make it exact. The integrating factor μ is a function of x and y that when multiplied to the entire equation, makes it exact. It should satisfy the following condition:

(μ * ∂/∂x) (2xsin(y) + 3x^2) = (μ * ∂/∂y) (x^2cos(y) + 1/y)

Let's find μ:

(μ * (2sin(y) + 6x))dx = (μ * (-x^2sin(y) - 1/y^2))dy

Now, we equate the coefficients of dx and dy:

μ * (2sin(y) + 6x) = μ * (-x^2sin(y) - 1/y^2)

μ * (2sin(y) + 6x) = -μx^2sin(y) - μ/y^2

We can separate the terms with μ and those without μ:

2μsin(y) + 6μx = -μx^2sin(y) - μ/y^2

Now, we'll equate the coefficients of like terms on both sides of the equation:

2μsin(y) = -μx^2sin(y) 6μx = -μ/y^2

Now, we solve for μ:

From the first equation: 2μsin(y) = -μx^2sin(y)

Divide both sides by sin(y):

2μ = -μx^2

Now, we can isolate μ:

2μ = -μx^2 2μ/μ = -x^2

2 = -x^2

x^2 = -2

x = ±√(-2)

However, we're working with real numbers, and the square root of a negative number is not real. Therefore, there is no real-valued integrating factor μ, and this equation is not exact.

This means that it cannot be solved using the method of exact differentials, and alternative methods, such as separation of variables or integrating factors, may be needed.