Bernoulli equation

Definition

$y^{\prime }+p(x)y=g(x)y^a$

The equation is linear if $a=0,1$ and non-linear if $a\ne 0,1$

Solution

We can reduce it to a Linear ODE by substituting $u=y^{1-a}$ Then, $$\begin{gathered} u^{\prime }=(1-a)y^{-a}{y}^{\prime }=(1-a)y^{-a}(g{y}^a-py)=(1-a)(g-p{y}^{1-a})=(1-a)(g-pu) \\ \Rightarrow u^{\prime }+(1-a)pu=(1-a)g \end{gathered}$$ Since the form of the equation is Non-Homegeneous Linear ODE, we can find $u=y^{1-\alpha }$

Examples

&y' + xy = \frac{x}{y}\\ &\text{substitute}\ u:=y^{2} \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{u}}\frac{du}{dx}\\ &\frac{1}{2\sqrt{u}}\frac{du}{dx} + x\sqrt{u} = \frac{x}{\sqrt{u}} \Rightarrow \frac{du}{dx} + 2xu = 2x\\ &\text{By the solution of separable ODE}\\ &u = 1-\exp(-x^{2})C\\ &\Rightarrow y^{2} = 1-\exp(-x^{2})C \end{aligned}$$