<s>[INST] How do I solve differential equation $\frac{d}{dt}x(t)=(2x(t)+8)(t^4+2t^2-t)$? How do I solve differential equation $\frac{dx}{dt}=(2x+8)(t^4+2t^2-t)$?
$\frac{dx}{2x(t)+8}=dt(t^4+2t^2-t)$
$\frac{1}{2}\ln(2x+8)=\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2}$
$e^{\ln(2x+8)}=e^{2(\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2})}$
$2x+8=e^{2(\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2})}$
$2x=e^{2(\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2})}-8$
$x=\frac{e}{2}^{2(\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2})}-4$
$x(0)=3$
Is the above calculation correct?
I can put that values, so I get:
$\frac{1}{2}\ln(14)=-C$
Where do I write value of the constant C in the solution $x(t)$?
 [/INST] $\frac{1}{2}\ln(2x+8)=\frac{t^5}{5}+2\frac{t^3}{3}-\frac{t^2}{2}+C$
I advocate waiting until the end to solve for $C.$  There is a lot of algebra to follow, and it is easier if there is one less thing to think about.
For example,  $C$ is an arbitrary constant.  Multiply by $2, 2C$ is equally arbitrary, so just leave it as $C.$
$\ln(2x+8)=2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2}+C$
$e^{\ln(2x+8)} = e^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2}+C)}\\
2x+8 = e^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}e^C\\
2x+8 = Ce^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}$
Same game, $e^C$ and $C$ are equally arbitrary constants.
$x+4 = Ce^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}\\
x = Ce^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}-4$
Now, lets tackle $C$ in the initial condition.
$x(0) = C - 4 = 3$
$x = 7e^{(2\frac{t^5}{5}+4\frac{t^3}{3}-{t^2})}-4$
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