I'm trying to solve this homework question but the two variables is throwing me off. Which one is my standard $t$? How do I handle both variables? I'm to solve this Bernoulli equation via substitution with $v=y^{1-n}$ if $n\ne0, 1$. The textbook references Leibniz found this method to reduce a Bernoulli equation down to a linear equation in 1696.
The actual question is:
$$
y'=\epsilon y-\theta y^3,\space\space\space \epsilon > 0, \space\space\space\theta > 0
$$
and mentions that this equation occurs in the study of the stability of fluid flow.
I get to the point where I have $y'=-\frac{v'y^3}{2}$ and substitute that back in and get:
$$
v'+2v\epsilon=2\theta
$$
but I'm not sure how to get an integrating factor from this point. If typically $y'+p(t)y=g(t)y^n$, am I using $p(\epsilon)$ or $g(\theta)$ when finding the integrating factor?
Best Answer
$y'=\epsilon y-\theta y^3$ is a separable ODE. Just integrate $dx=\frac{dy}{\epsilon y-\theta y^3}$ to solve it.
Considering it as a Bernoulli ODE will finally lead to the same integral. But you can do it anyway.
$v'+2v\epsilon=2\theta$ is a classical first order linear ODE
The solution of the related homogeneous ODE $v'+2v\epsilon=0$ is $v=c e^{-2\epsilon x}$
So, the change of function $v(x)=f(x)e^{-2\epsilon x}$ into $v'+2v\epsilon=2\theta$ will lead to a separable ODE of the simplest kind.