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Differential equations with separable variables Step-by-Step

For example, you have entered (calculator here):
2xy(x)+(x−1)ddxy(x)=02 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−1x - 1
We get the equation:
2xy(x)+(x−1)ddxy(x)x−1=0\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(x)=2xx−1P{\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

dyy=−P(x)dx\frac{dy}{y} = - P{\left(x \right)} dx, if y is not equal to 0
∫1y dy=−∫P(x) dx\int \frac{1}{y}\, dy = - \int P{\left(x \right)}\, dx
log⁡(∣y∣)=−∫P(x) dx\log{\left(\left|{y}\right| \right)} = - \int P{\left(x \right)}\, dx
Or,
∣y∣=e−∫P(x) dx\left|{y}\right| = e^{- \int P{\left(x \right)}\, dx}
Therefore,
y1=e−∫P(x) dxy_{1} = e^{- \int P{\left(x \right)}\, dx}
y2=−e−∫P(x) dxy_{2} = - e^{- \int P{\left(x \right)}\, dx}
The expression indicates that it is necessary to find the integral:
∫P(x) dx\int P{\left(x \right)}\, dx
Because
P(x)=2xx−1P{\left(x \right)} = \frac{2 x}{x - 1}, then
∫P(x) dx\int P{\left(x \right)}\, dx =
= ∫2xx−1 dx=(2x+2log⁡(x−1))+Const\int \frac{2 x}{x - 1}\, dx = \left(2 x + 2 \log{\left(x - 1 \right)}\right) + Const
Detailed solution of the integral
So, solution of the homogeneous linear equation:
y1=eC1−2x(x−1)2y_{1} = \frac{e^{C_{1} - 2 x}}{\left(x - 1\right)^{2}}
y2=−eC2−2x(x−1)2y_{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=Ce−2x(x−1)2y = \frac{C e^{- 2 x}}{\left(x - 1\right)^{2}}
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