Question

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

Answer 1

Ответ: x²*y²+x²/y²=C.

Пошаговое объяснение:

Запишем равнение в виде P*dx+Q*dy=0, где P=x*y²+x/y² и Q=x²*y-x²/y³. Найдём частные производные dP/dy и dQ/dx: dP/dy=2*x*y-2*x/y³, dQ/dx=2*x*y-2*x/y³. Так как dP/dy=dQ/dx, то левая часть данного уравнения представляет собой полный дифференциал du некоторой функции u(x,y). Отсюда P=du/dx=x*y²+x/y² и тогда u=∫(x*y²+x/y²)*dx=1/2*x²*y²+1/2*x²/y²+f(y), где f(y) - неизвестная пока функция от y. Дифференцируя это выражение по y и приравнивая к Q, приходим к уравнению x²*y-x²/y³+f'(y)=x²*y-x²/y³. Отсюда f'(y)=0 и тогда f(y)=C1, где C1 - произвольная постоянная. И так как du=0, то u=const и поэтому u(x,y)=1/2*x²*y²+1/2*x²/y²+C1=C2, где C2 - также произвольная постоянная. Отсюда, полагая C2-C1=C3, получаем  1/2*x²*y²+1/2*x²/y²=С3, или, полагая 2*С3=С, x²*y²+x²/y²=C.

Answer 2

The given equation is a first-order ordinary differential equation that is not exact in its current form. An exact equation is one where the partial derivatives of the coefficients of dx and dy with respect to the other variable are equal.

Let's examine the given equation:

$xy^2 + \frac{x}{y^2}$dx + $x^2y - \frac{x^2}{y^3}$dy = 0.

We have: M(x, y) = xy^2 + \frac{x}{y^2}, N(x, y) = x^2y - \frac{x^2}{y^3}.

Now, let's compute the partial derivatives of M with respect to y and N with respect to x:

∂M/∂y = 2xy - \frac{2x}{y^3}, ∂N/∂x = 2xy - \frac{2x}{y^3}.

We can see that the partial derivatives are equal, so the equation is almost exact. To make it exact, we need to multiply the entire equation by an integrating factor. The integrating factor is typically found by taking the ratio of the coefficient of dy in the equation to ∂M/∂y or ∂N/∂x. In this case, let's use the coefficient of dy in the equation:

$x^2y - \frac{x^2}{y^3}$.

The integrating factor, μ(x, y), is given by:

μ(x, y) = e^(∫(∂N/∂x - ∂M/∂y) dx).

Substitute the values of ∂N/∂x and ∂M/∂y into the above formula:

μ(x, y) = e^(∫(2xy - \frac{2x}{y^3} - 2xy + \frac{2x}{y^3}) dx) = e^(∫0 dx) = e^0 = 1.

Since the integrating factor is 1, you don't need to multiply the equation by anything, and the equation remains the same:

xy^2 + \frac{x}{y^2} dx + x^2y - \frac{x^2}{y^3} dy = 0.

This equation is now exact and can be solved using the method of exact differential equations. To find the solution, you need to integrate M with respect to x and N with respect to y, and then find the function φ(x, y) such that:

∂φ/∂x = M(x, y), ∂φ/∂y = N(x, y).

The solution will be φ(x, y) = constant.