Nonlinear ODE $y''+y'+\frac {1}{y}=0$, $\int yy''dy$ and $\int yy''dt$ $$y''(t)+y'(t)+\frac {1}{y(t)}=0$$
(by y' and y'' I mean $\frac{dy}{dt}$ and $\frac{d^2y}{dt^2}$)
First, I only know basic-ish Calculus but I'm willing to do some reading. Second, I'd appreciate some help with my (extra?) doubts (marked with "‡"). Third, this and this are kinda similar and I understand how they work but I want the function that multiplies $y''$ to be the same (or a multiple but unless that makes it easier, let's stick with this) as the one that multiplies $y'$. A square in the first derivative $y'$ actually makes sense for what I was trying to do but I'm not sure if it'll simplify things or if these are helpful here:$$(yy')'=yy''+y'y'$$ $$(y'/y)=\frac{y''}y-\frac{y'^2}{y^2}$$
My failed attempt(s)---
I know that not all nonlinear DEs are [nicely?] solvable but I just wanna make sure. My issue with this one is that if we integrate both sides in function of $t$ we get $$\int \frac{d}{dt}\left(\frac{dy}{dt}\right) dt+\int\frac{dy}{dt}dt+\int\frac {1}{y}dt=0\Rightarrow$$
$$y'+y+\int\frac {1}{y}dt=0$$ and I can't get rid of the last bit ‡[I've heard that you should be cautious with simplifying differentials like dt and dy algebraically but I don't understand why. Can you give me some examples?]. In function of y:
$$\int \frac{d}{dt}\left(\frac{dy}{dt}\right) dy+\int\frac{dy}{dt}dy+\int\frac {1}{y}dy=0\Rightarrow$$
$$\int \frac{d}{dt}\left(\frac{dy}{dt}\right) dy+\int\frac{dy}{dt}dy+\ln|y|=0$$ and I can't get rid of the first two bits.
If we multiply the equation by $y$:
$$yy''+yy'+1=0$$
Now, to successfully integrate in function of something we need to be able to solve either $$\int yy''+yy' dt$$ or $$\int yy''+yy' dy$$. Let's try in function of $t$ first.
$$\int yy' dt=\int y\frac{dy}{dt}dt=\frac {y^2}{2}+C$$, so there's that.
(integrating by parts with the integral I want on the left side and yes, it might be weird but I think it's so much simpler)
$$\int yy'' dt=yy'-\int y'y' dt$$ or $$\int yy'' dt=y''\int y dt-\int (y'''\int y dt)dt$$
$\int y'y'dt=y'y-\int y''y dt$ Substituting this into the nicer one above:
$$\int yy'' dt=yy'-\bigg(y'y-\int y''y dt\bigg)$$, which cancels out and makes me sad
Trying to integrate by parts like this also doesn't work:$\int yy''*1 dt$.
Let's try in function of $y$.
‡[Now, for the stuff above, Wolfram Alpha didn't spit out anything useful but it gave me some results for the integrations by parts below I don't think are right. Is my boi WA wrong?]
$$\int yy' dy=y\int \frac{dy}{dt}dy-\int 1(\int \frac{dy}{dt}dy)dy$$ or
$$\int yy' dy=y'\frac{y^2}{2}-\int \frac{d}{dy}\frac{dy}{dt}\frac{y^2}{2}dy=y'\frac{y^2}{2}-\int \frac{y^2}{2}d\frac{dy}{dt}$$
$$\int \frac{y^2}{2}*1d\frac{dy}{dt}=\frac{y^2}{2}y'-\int \frac{d(y^2/2)}{dy'}y'dy'=\frac{y^2}{2}y'-\int yy' dy$$ Substituting above: $$\int yy' dy=y'\frac{y^2}{2}-(\frac{y^2}{2}y'-\int yy' dy)$$, which leads us to the stunning conclusion that $0=0$.
Nevertheless, Wolfram Alpha says that $\int yy' dy=\frac{y^2}{2}y'+C$ but if $y=t^3$, $$\int yy' dy=\int t^3(3t^2)d(t^3)=\int t^3(3t^2)(3t^2)dt=\frac{9t^8}{8}+C\neq\frac{3t^8}{2}$$. Kinda close, though? It also says that $$\int yy'' dy=\frac{y^2}{2}y''+C$$ . Using that info to try to solve our equation that's multiplied by $y$, we can only further complicate the problem.
I've tried doing $u=\frac{1}{y}$ and $u=x'$ but those didn't seem to lead anywhere.
[PS: Also, this doesn't come from a textbook or something, I just thought about a Physics thing (which I'm not even sure is reasonable) and got stuck on the math part. I've become more interested in the mathematics behind it than on whether or not it's useful, so the original situation doesn't matter anymore. But do see the comments if you're curious.
PPS: Formatting math expressions is a thing that rhymes with "itch". But thanks for reading!]
 A: Assume $y=f(x)$ is the solution. We can also work with inverse functions. Hence let $x=f^{(-1)}(y)$ is the solution. Then
$$
\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{\dot{x}}\tag 1
$$
and
$$
\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{\dot{x}}\right)=\frac{d}{dy}\left(\frac{1}{\dot{x}}\right)\frac{dy}{dx}=\frac{d}{dy}\left(\frac{1}{\dot{x}}\right)\frac{1}{\dot{x}}\tag 2
$$
Hence equation
$$
y''+y'+\frac{1}{y}=0\tag 3
$$
becomes (setting $\dot{x}=1/G(y)$)
$$
G'(y)G(y)+G(y)+\frac{1}{y}=0\Leftrightarrow 
$$
$$
-iy=\frac{2\exp\left(-\frac{(y+G(y))^2}{2}\right)}{C_1-i\sqrt{2\pi}\cdot\textrm{erf}\left(\frac{y+G(y)}{\sqrt{2}}\right)}
$$
Assume now the equation
$$
Y=\frac{2\exp\left(-\frac{1}{2}X^2\right)}{C_1-i\sqrt{2\pi}\cdot\textrm{erf}\left(\frac{1}{\sqrt{2}}X\right)}\tag 4
$$
Have solution $X=L(Y)$. Then
$$
y+G(y)=L(-iy)\Leftrightarrow y+\frac{1}{\dot{x}}=L(-iy)\Leftrightarrow
$$
$$
x=\int^{y(x)}_{C_0}\frac{dt}{L(-it)-t}\tag 5
$$
Note. In equations of the form $y''=f(y,y')$, the transformation of variables (1),(2) always reduces the order of the equation by 1.
