I'm trying to solve an ODE of the form \begin{equation} x\frac{dx}{dt}+\sqrt{x^4-x^2+x+1}=0. \end{equation} I'm unable to solve this by just integrating, so I've been trying all sorts of different methods. I don't expect anyone to know how to solve this, (there's a good chance there's no solution consisting of elementary function), just to point out what false assumption or mistake I've made in this attempt!
I look for a function f(x) such that
\begin{equation} e^tf(x)x\frac{dx}{dt}+e^tf(x)\sqrt{x^4-x^2+x+1}=\frac{d}{dt}e^tf(x)\sqrt{x^4-x^2+x+1}. \end{equation} I'll use g(x) for the sake of brevity here. Write out the triple product: \begin{equation} e^tf(x)x\frac{dx}{dt}+e^tf(x)g(x)=\frac{d}{dt}e^tf(x)g(x)=e^t\left[f(x)g(x)+\frac{df}{dx}\frac{dx}{dt}g(x)+f(x)\frac{dg}{dx}\frac{dx}{dt}\right], \end{equation} cancel out the $e^t$ and $f(x)g(x)$, \begin{equation} f(x)x\frac{dx}{dt}=\frac{df}{dx}\frac{dx}{dt}g(x)+f(x)\frac{dg}{dx}\frac{dx}{dt}, \end{equation} cancel out $dx/dt$ \begin{equation} f(x)x=\frac{df}{dx}g(x)+f(x)\frac{dg}{dx}, \end{equation} rewrite \begin{equation} \frac{1}{f(x)}\frac{df}{dx}=\frac{x}{g(x)}-\frac{1}{g(x)}\frac{dg}{dx}, \end{equation} integrate \begin{equation} \text{ln}(f(x))=\text{ln}(\frac{1}{g(x)})+\int\frac{x}{g(x)}dx\\ f(x)=\frac{1}{g(x)}\text{exp}(\int\frac{x}{g(x)}dx). \end{equation} So my ODE becomes \begin{equation} \frac{d}{dt}\text{exp}\left[t+\int\frac{x}{g(x)}dx\right]=0, \end{equation} integrate and take the natural logarithm, \begin{equation} \text{exp}\left[t+\int\frac{x}{g(x)}dx\right]=c\\ t+\int\frac{x}{g(x)}dx=c, \end{equation} I believe I can take a derivative here, as $dx=\frac{dx}{dt}dt$, \begin{equation} 1+\frac{x}{g(x)}=1+\frac{x}{\sqrt{x^4-x^2+x+1}}=0. \end{equation} This is just a quartic equation which could be solved given some google searching and following the procedure to solve it. My problem is that the ODE describes a dynamic system which I know is not described by a static solution, so why do I get such a result? I appreciate if you've taken the time to read this.
Writing $$t'=-\frac x {\sqrt{(x+1)(x^3-x^2+1)}}$$ the real root of the cubic is $$a=\frac{1}{3} \left(1-2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{25}{2}\right)\right)\right)$$ which gives $$t'=-\frac x {\sqrt{(x+1)(x-a)(x-b)(x-c)}}$$ where $b$ and $c$ are the two remaining complex conjugate roots of the cubic. For sure, looking at the rhs, we can expect some complicated elliptic integrals.
This would give $$t+C=-\frac{2 \sqrt{a+1}}{\sqrt{(c+1)(b-a)} }\Big[ \frac {T_1}{\sqrt{a+1}}-T_2\sqrt{-(a+1) }\Big]$$ where $$T_1=F\left(\sin ^{-1}\left(\sqrt{\frac{(c+1) (a-x)}{(x+1) (a-c)}}\right)|\frac{(b+1) (a-c)}{(c+1) (a-b)}\right)$$ $$T_2=\Pi \left(\frac{c-a}{c+1};\sin ^{-1}\left(\sqrt{\frac{(c+1) (a-x)}{(x+1) (a-c)}}\right)|\frac{(b+1) (a-c)}{(c+1) (a-b)}\right)$$
I hope and wish no mistakes in the simplifications of the monster produced by a CAS.
Now, have fun !