Question

(y²-2xy)dx+x²dy=0 диф. уравнение

Answer 1

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

Разделим на $dx$:

$y^2-2xy+x^2\cdot\dfrac{dy}{dx}= 0$

Разделим на $x^2$:

$\dfrac{y^2}{x^2} -2\cdot\dfrac{y}{x} +\dfrac{dy}{dx}= 0$

$\left(\dfrac{y}{x}\right)^2 -2\cdot\dfrac{y}{x} +y'= 0$

Замена:

$\dfrac{y}{x} =t$

$\Rightarrow y=tx$

$\Rightarrow y'=t'x+tx'=t'x+t$

Получим уравнение:

$t^2 -2t +t'x+t= 0$

$t^2 -t +t'x= 0$

$t'x= t-t^2$

$x\cdot\dfrac{dt}{dx} = t-t^2$

$\dfrac{dt}{ t-t^2} =\dfrac{dx}{x}$

$\int\dfrac{dt}{ t-t^2} =\int\dfrac{dx}{x}$

Чтобы найти интеграл, стоящий в левой части, подынтегральную дробь представим в виде суммы составляющих:

$\dfrac{1}{ t-t^2}=\dfrac{A}{t} +\dfrac{B}{1-t} =\dfrac{A(1-t)+Bt}{t(1-t)} =\dfrac{A-At+Bt}{t(1-t)} =\dfrac{(B-A)T+A}{t(1-t)}$

Дроби равны, знаменатели равны. Значит, равны и числители:

$1=(B-A)t+A$

Условие равенства:

$\begin{cases} A=1 \\ B-A=0 \end{cases}$

$\begin{cases} A=1 \\ B=A=1 \end{cases}$

Значит, дробь раскладывается на составляющие следующим образом:

$\dfrac{1}{ t-t^2}=\dfrac{1}{t} +\dfrac{1}{1-t}$

Возвращаемся к интегрированию:

$\int\dfrac{dt}{ t-t^2} =\int\dfrac{dx}{x}$

$\int\dfrac{dt}{ t}+\int\dfrac{dt}{1-t} =\int\dfrac{dx}{x}$

Для второй дроби выполним подведение под знак дифференциала:

$\int\dfrac{dt}{ t}-\int\dfrac{d(1-t)}{1-t} =\int\dfrac{dx}{x}$

Интегрируем:

$\ln|t|-\ln|1-t|=\ln|x|-\ln|C|$

$\ln\left|\dfrac{t}{1-t}\right|=\ln\left|\dfrac{x}{C}\right|$

$\dfrac{t}{1-t}=\dfrac{x}{C}$

Обратная замена:

$\dfrac{\frac{y}{x} }{1-\frac{y}{x} }=\dfrac{x}{C}$

$\dfrac{y }{x-y }=\dfrac{x}{C}$

$\dfrac{x-y }{y }=\dfrac{C}{x}$

$\dfrac{x }{y }-1=\dfrac{C}{x}$

$\dfrac{x }{y }=1+\dfrac{C}{x}$

$\dfrac{x }{y }=\dfrac{x+C}{x}$

$\boxed{y=\dfrac{x^2}{x+C}}$

Answer 2

The given differential equation is:

(y² - 2xy)dx + x²dy = 0

To solve this differential equation, we can check if it is exact or not. A differential equation is exact if it can be written in the form M(x, y)dx + N(x, y)dy = 0, and the following condition is satisfied:

∂M/∂y = ∂N/∂x

where ∂M/∂y represents the partial derivative of M with respect to y, and ∂N/∂x represents the partial derivative of N with respect to x.

Let's check if the given equation is exact:

M(x, y) = y² - 2xy N(x, y) = x²

Now, calculate the partial derivatives:

∂M/∂y = 2y - 2x ∂N/∂x = 2x

The equation is exact if ∂M/∂y = ∂N/∂x. Let's check:

2y - 2x = 2x

This equation is not true; hence, the given differential equation is not exact.

To solve the differential equation, we can try to make it exact by using an integrating factor. An integrating factor is a function that multiplies both sides of the equation to make it exact.

To find the integrating factor (IF), we use the following formula:

IF = e^(∫(∂N/∂x - ∂M/∂y) / N)dx

In our case, N(x, y) = x², and (∂N/∂x - ∂M/∂y) = 2x - (2y - 2x) = 4x - 2y.

So the integrating factor is:

IF = e^(∫(4x - 2y) / x²)dx

Now, we can find the integral:

∫(4x - 2y) / x² dx = ∫(4 - 2y/x) dx = 4x - 2∫(y/x) dx = 4x - 2∫(y/x) dx = 4x - 2∫(y * x^(-1)) dx = 4x - 2∫(y * x^(-1)) dx = 4x - 2∫(y * x^(-1)) dx = 4x - 2∫y * d(x^(-1)) = 4x - 2y * x^(-1) + C

where C is the constant of integration.

Therefore, the integrating factor is e^(4x - 2y * x^(-1) + C).

Now, multiply both sides of the original differential equation by the integrating factor:

e^(4x - 2y * x^(-1) + C) * ((y² - 2xy)dx + x²dy) = 0

After multiplying, we need to check if the equation is now exact. If it is, we can proceed to solve it further. However, if it is still not exact, we would need to try different methods of solving the equation.