Dtyjlui

I have to transform a second order differential equation into a Bernoulli type differential equation, however, I am having some trouble.
The original equation is:
begin{equation}
ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0
end{equation}
By defining a new function:
begin{equation}
phi = frac{dt}{du}
end{equation}
The original equation has to be transformed into:
begin{equation}
frac{dphi}{du} + frac{1}{u}phi = (y-x_{0}beta u) phi^{2}
end{equation}
Would anyone be willing to suggest any ideas for how this might be achieved?
Thank you!

We have

$$phi=frac{dt}{du}implies frac{dphi}{du}=frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}=-phifrac{frac{d^2u}{dt^2}}{left(frac{du}{dt}right)^2}=-phi^3frac{d^2u}{dt^2}$$

Then the original differential equation transforms from

$$ufrac{d^{2}u}{dt^{2}} - bigg(frac{du}{dt} bigg)^{2} +(gamma -x_{0}beta u)ufrac{du}{dt}=0$$

to

$$-frac u{phi^3}frac{dphi}{du}-frac1{phi^2}+(gamma-x_0beta u)frac u{phi}=0$$

Multiplying through by $-frac{phi^3}u$, one gets to the required answer.

$$frac{dphi}{du}+frac1uphi-(gamma-x_0beta u)phi^2=0$$

(I think you have a couple of typos in the question)

Same solution, but with more details concerning the computations: Leibniz notation $frac{d}{dx}$ is very concise but you can lose track of what you really do, especially for high order derivatives.

$u$ is some function of $t$, $u=f(t)$. Let's denote $f^{-1}$ the reciprocal function (it must be assumed that $u$ is monotonic on a neighborhood of $t$ to make the change of function meaningful):
$$
fcirc f^{-1}(u)=u
$$
Using the chain rule we get:
$$
frac{d}{du}(fcirc f^{-1}(u))=f'(f^{-1}(u))(f^{-1})'(u)=1
$$
thus, by identification:
$$
phi(u)triangleq (f^{-1})'(u)=frac{1}{f'(f^{-1}(u))}triangleqfrac{dt}{du}
$$
which can be rewritten as (in a more concise form):
$$
frac{du}{dt}=frac{1}{phi}
$$
Now if you derivate once more:
begin{align}
frac{d}{du}phi(u)&=frac{d}{du}frac{1}{f'(f^{-1}(u))}\
&=-frac{(f^{-1})'(u)f''(f^{-1}(u))}{(f'(f^{-1}(u)))^2}\
&=-phi^3(u)f''(f^{-1}(u))
end{align}
which can be rewritten as (in a more concise form):
$$
frac{d^2u}{dt^2}=-frac{1}{phi^3}frac{dphi}{du}
$$
Now you can do direct substitution into your initial equation:
$$
ufrac{d^2u}{dt^2}−(frac{du}{dt})^2+(γ−x_0βu)ufrac{du}{dt}=0
$$
hence
$$
-frac{u}{phi^3}frac{dphi}{du}-(frac{1}{phi})^2+(γ−x_0βu)ufrac{1}{phi}=0
$$
If you multiply by $-frac{phi^3}{u}$ you get your expected result:
$$
frac{dphi}{du}+frac{phi}{u}=(γ−x_0βu)phi^2
$$

Extra comment concerning (critics of the) Leibniz notation:

I personally do not like Leibniz notation for high order derivatives. By example, it is clear that the chain rule can be written as:
$$
frac{dy}{dt}=frac{dy}{dx}frac{dx}{dt}
$$
however for higher order derivatives great care must be taken, as this notation can also be misleading:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}frac{dx^2}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2
$$
which is wrong, the right formula is:
$$
frac{d^2y}{dt^2}=frac{d^2y}{dx^2}left(frac{dx}{dt}right)^2+frac{dy}{dx}frac{d^2x}{dt^2}
$$
You have not this problem with Lagrange notation:
$$
y(x(t))''=(y'(x(t))x'(t))'=y''(x(t))(x'(t))^2+y'(x(t))x''(t)
$$

So, personally I always have doubts when I write expressions like:
$$
frac{d}{du}frac{dt}{du}=frac{dt}{du}frac d{dt}frac{1}{frac{du}{dt}}
$$
(again it is a matter of personal habits, I do not say it is a bad thing to do that (see post comments), it is just that personally I don't feel confident when I write these kind of expressions).

This problem (and others) with the Leibniz notation is well known when teaching differential calculus, see:

H. Poincaré, La Notation Différentielle et l'enseignement (pdf)

J. Hadamard, La notion de différentielle dans l'enseignement (pdf)

unfortunately both in French, however you can find an English translation of Hadamard's article here.

You can also see:

Differentials, higher-order differentials and the derivative in the Leibnizian calculus (pdf)