when doing second order differential equation sometimes i suppose $y'=p$
$$\frac{d^2y}{dx^2}=\frac{dp}{dx}=\frac{dp}{dy}.\frac{dy}{dx}=p.\frac{dp}{dy}$$
what is the relation of chain rule with second order differential?
$\displaystyle \frac{u"}{u'}+(1-\frac{1}{x-1})=0 $
I want to integrate this. Should i suppose $u'=b$ and $u"=b'$ ?
Or should I use $\displaystyle\frac{d^2y}{dx^2}=\frac{dy}{dx}=\frac{dp}{dy}.\frac{dy}{dx}=p.\frac{dp}{dy}$ ??
Here you have the case where $u$ (dependent variable) is missing. With the substitution $u'=y$ you get a first order linear differential equation:
$$y^{'}+\left(1-\frac1x\right)y=0$$
This first order differential equation can be solved by separating the variables.
If the independent variable is missing then you use
$$\frac{d^2y}{dx^2}=\frac{dp}{dx}=\frac{dp}{dy}.\frac{dy}{dx}=p.\frac{dp}{dy}$$
Example:
$\frac{d^2y}{dx^2}=-\frac{2}{1-y}\left(\frac{dy}{dx}\right)^2$
$\frac{dy}{dx}=p \quad (\color{blue}*)$
$p\frac{dp}{dy}=-\frac{2}{1-y} \cdot p^2$
Dividing the equation by $p^2$ and multipling it by $dy$
$\frac{dp}{p}=-\frac{2}{1-y} \ dy$
$p=c_1\cdot (1-y)^2$
We can use $*$ and get $\frac{dy}{dx}=c_1\cdot (1-y)^2$.
After the variables has been separated this differential equation can be solved as well.