# 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:

$$y = \frac{C e^{- 2 x}}{\left(x - 1\right)^{2}}$$