Question

X(x-1)y'+2xy=1 pleaseeeeee

Answer 1

$x(x - 1)y' + 2xy = 1$

Разделим обе части уравнения на $x(x - 1) \neq 0$

$y' + \dfrac{2}{x - 1}y = \dfrac{1}{x(x - 1)}$

Имеем линейное дифференциальное уравнение первого порядка

Воспользуемся методом Бернулли: $y = uv; \ y' = u'v+uv'$, где $u = u(x), \ v = v(x)$

Имеем:

$u'v + uv' + \dfrac{2}{x - 1}uv = \dfrac{1}{x(x - 1)}$

$u'v + u\left(v' + \dfrac{2}{x - 1}v\right) = \dfrac{1}{x(x - 1)}$

Пусть $v' + \dfrac{2}{x - 1}v = 0$. Тогда $u'v = \dfrac{1}{x(x-1)}$

Решим первое дифференциальное уравнение:

$v' + \dfrac{2}{x - 1}v = 0$

$\dfrac{dv}{dx} =- \dfrac{2}{x - 1}v$

$\dfrac{dv}{v} = -\dfrac{2}{x-1}dx$

$\displaystyle \int\dfrac{dv}{v} = -\int \dfrac{2}{x-1}dx$

$\ln|v| = - 2\ln |x - 1|$

$\ln|v| = \ln \dfrac{1}{(x-1)^{2}}$

$v = \dfrac{1}{(x - 1)^{2}}$

Решим второе дифференциальное уравнение:

$u'v = \dfrac{1}{x(x-1)}$

$\dfrac{du}{dx} \cdot \dfrac{1}{(x - 1)^{2}}= \dfrac{1}{x(x-1)}$

$du = \dfrac{x-1}{x} dx$

$du =\left(1 - \dfrac{1}{x}\right) dx$

$\displaystyle \int du = \int \left(1 - \dfrac{1}{x}\right) dx$

$u = x - \ln|x| + C$

Обратная замена:

$y = uv = (x - \ln|x| + C) \cdot \dfrac{1}{(x - 1)^{2}} = \dfrac{x - \ln |x| + C}{(x-1)^{2}}$

Ответ: $y = \dfrac{x - \ln |x| + C}{(x-1)^{2}}$

Answer 2

It looks like you've provided a first-order linear ordinary differential equation. The equation you've given is:

X(x-1)y' + 2xy = 1

To solve this differential equation, we can use the method of integrating factors. The general form of a first-order linear ODE is:

dy/dx + P(x)y = Q(x)

In your case, P(x) = 2x/(x-1) and Q(x) = 1/X. We'll first rewrite the equation in the standard form:

y' + (2x/(x-1))y = 1/X

Now, to find the integrating factor (IF), we multiply the coefficient of y (which is P(x)) by e^(integral of P(x) dx):

IF = e^(∫(2x/(x-1)) dx)

Next, we'll integrate P(x):

∫(2x/(x-1)) dx = 2∫(1 + 1/(x-1)) dx = 2x + 2ln|x-1| + C

So, the integrating factor is:

IF = e^(2x + 2ln|x-1| + C) = e^(2x) * e^(2ln|x-1|) * e^C = e^(2x) * |x-1| * e^C

Now we multiply both sides of the equation by the integrating factor:

e^(2x) * |x-1| * y' + 2e^(2x) * |x-1| * y = e^(2x) * |x-1| * (1/X)

Notice that we have the left side in the form of the derivative of a product, (e^(2x) * |x-1| * y). Applying the product rule, we can simplify the left side:

d(e^(2x) * |x-1| * y)/dx = e^(2x) * |x-1| * (1/X)

Now we integrate both sides with respect to x:

∫ d(e^(2x) * |x-1| * y)/dx dx = ∫ e^(2x) * |x-1| * (1/X) dx

This gives us:

e^(2x) * |x-1| * y = ∫ e^(2x) * |x-1| * (1/X) dx + C

At this point, the integral on the right side may not have a simple closed-form solution, so we might need to leave it in this form or use numerical methods to approximate it.

The solution to the original differential equation is:

y = (1/(e^(2x) * |x-1|)) * (∫ e^(2x) * |x-1| * (1/X) dx + C)

Please note that this solution might involve an implicit form due to the integral, and further analysis or approximation may be needed depending on the specific context of the problem.