Differential equations with separable variables Step-by-Step
$$2 x y{\left(x \right)} + \left(x - 1\right) \frac{d}{d x} y{\left(x \right)} = 0$$
Detail solution
Divide both sides of the equation by the multiplier of the derivative of y':$$x - 1$$
We get the equation:
$$\frac{2 x y{\left(x \right)} + \left(x - 1\right) \frac{d}{d x} y{\left(x \right)}}{x - 1} = 0$$
This differential equation has the form:
y' + P(x)y = 0,
where
$$P{\left(x \right)} = \frac{2 x}{x - 1}$$
and
and it is called linear homogeneous
differential first-order equation:
It's an equation with multiple variables.
The equation is solved using following steps:
From y' + P(x)y = 0 you get
$$\frac{dy}{y} = - P{\left(x \right)} dx$$, if y is not equal to 0
$$\int \frac{1}{y}\, dy = - \int P{\left(x \right)}\, dx$$
$$\log{\left(\left|{y}\right| \right)} = - \int P{\left(x \right)}\, dx$$
Or,
$$\left|{y}\right| = e^{- \int P{\left(x \right)}\, dx}$$
Therefore,
$$y_{1} = e^{- \int P{\left(x \right)}\, dx}$$
$$y_{2} = - e^{- \int P{\left(x \right)}\, dx}$$
The expression indicates that it is necessary to find the integral:
$$\int P{\left(x \right)}\, dx$$
Because
$$P{\left(x \right)} = \frac{2 x}{x - 1}$$, then
$$\int P{\left(x \right)}\, dx$$ =
= $$\int \frac{2 x}{x - 1}\, dx = \left(2 x + 2 \log{\left(x - 1 \right)}\right) + Const$$
So, solution of the homogeneous linear equation:
$$y_{1} = \frac{e^{C_{1} - 2 x}}{\left(x - 1\right)^{2}}$$
$$y_{2} = - \frac{e^{C_{2} - 2 x}}{\left(x - 1\right)^{2}}$$
that leads to the correspondent solution
for any constant C, not equal to zero:
$$y = \frac{C e^{- 2 x}}{\left(x - 1\right)^{2}}$$