How to solve $\dfrac{\mathrm dx(t)}{\mathrm dt}=(2x(t)+6)(-2t^5+4t^4+2t^2)$ if $x(0)=-3$? How do I solve the following differential equation?
$$\frac{\mathrm dx(t)}{\mathrm dt}=(2x(t)+6)(-2t^5+4t^4+2t^2),\quad x(0)=-3$$
I've searched on Wolfram Alpha, and it shows: expand the $(2x(t)+6)(-2t^5+4t^4+2t^2)$, distribute $-2t^5+4t^4+2t^2$ over $6+2x(t)$. It's totally incomprehensible. 
Thanks.
 A: You start in a root of the right side, and it is a root independent of $t$. So noting is moving, zero velocity. The solution is the constant $-3$.
Use uniqueness to show that no other solution is possible.
A: $x(t)=-3$ is a singular solution of it while no other solution exists due to the initial condition $x(0)=-3$.
A: Simply, 
$$
\begin{aligned}
\dfrac{\mathrm{d}x(t)}{\mathrm{d}t}&=(2x(t)+6)(-2t^5+4t^4+2t^2)\\
\dfrac{\mathrm{d}x(t)}{2x(t)+6}&=(-2t^5+4t^4+2t^2)\mathrm{d}t\\
\end{aligned}
$$
Integrating both sides will give you the answer
$$
\begin{aligned}
\int \dfrac{\mathrm{d}x(t)}{2x(t)+6}&=\int(-2t^5+4t^4+2t^2)\mathrm{d}t\\
\frac12\log|2x(t)+6|&=-\frac13t^6+\frac45t^5+\frac23t^3+C
\end{aligned}
$$
The constant $C$ depends on your initial condition.
A: $\dfrac{dx}{dt}=(2x+6)(-2t^5+4t^4+2t^2)$
$\implies \displaystyle \int \dfrac{dx}{2x+6}=\int -2t^5+4t^4+2t^2\,dt$
$\implies \dfrac{1}{2}\ln|2x+6|=-\dfrac{1}{3}t^6+\dfrac{4}{5}t^5+\dfrac{2}{3}t^3+\ln C_1$
$\implies \ln|2x+6|=-\dfrac{2}{3}t^6+\dfrac{8}{5}t^5+\dfrac{4}{3}t^3+\ln C$
$\implies \ln\left|\dfrac{2x+6}{C}\right|=-\dfrac{2}{3}t^6+\dfrac{8}{5}t^5+\dfrac{4}{3}t^3$
$\implies 2x+6=Ce^{-\frac{2}{3}t^6+\frac{8}{5}t^5+\frac{4}{3}t^3}$
$\implies x=\dfrac{1}{2}[Ce^{-\frac{2}{3}t^6+\frac{8}{5}t^5+\frac{4}{3}t^3}-6]$
Using $x(0)=-3$
$\implies -3=\dfrac{1}{2}\left(C-6\right)$
$\implies C=0$
Hence the solution:
$x(t)=-3$
