Question

𝑦(2𝑥𝑦 + 1)𝑑𝑥 − 𝑥𝑑𝑦 = 0

Answer 1

Write in the form of a first order Bernoulli Ordinary Differential Equation

$$y'-\dfrac{1}{x}y=2y^2$$

The general solution is obtained by substituting

$$z=y^{1-n}=y^{1-2}=y^{-1}$$

Differentiating, we find:

$$z'=(y^{-1})'=-y^{-2}y'$$

Solve

$$y^{-2}y'-\dfrac{1}{x}y^{-1}=2$$

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

$$\mu(x)=e^{\int{1 \over x}dx}=e^{\ln x}=x$$

We can make sure that the function $x$ is the integrating factor

$$xz'+z=-2x$$

$$(xz)'=-2x$$

Integrate

$$\int d(xz)=-\int2xdx$$

$$xz=-x^2+C$$

$$xy^{-1}=-x^2+C$$

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