I've found an answer to a question, which is apparently incorrect but I have no idea why.
"Find the solution of the differential equation that satisfies the given initial condition."
$\frac{du}{dt}=\frac{2t^2+sec^2(t)}{2u}, u(0) = -3$
$\int2u.du = \int(2t+sec^2(t) )dt$
$\frac{2u^2}{2} = \frac{2t^2}{2}+tan(t) + C$
$u^2 = t^2+tan(t)+C$
$u = \sqrt{t^2+tan(t)+C}$
$u(0) = \sqrt{0^2+tan(0)+C}$ = -3
$=>\sqrt{C} = -3$
$=> C = 9$
So the solution is $u = \sqrt{t^2+tan(t)+9}$, yes?
Apparently my answer is wrong but I can't for the life of me see how.
Any indication of my mistake would be much appreciated?
Your mistake was that you didn't put in a $\pm$ in front of the square root. From there the solution should've gone as
$u = \pm\sqrt{t^2+\tan(t)+C}$
$u(0) = \pm\sqrt{0^2+\tan(0)+C}$ = -3
$\implies -\sqrt{C} = -3$
$\implies C = 9$
$u = -\sqrt{t^2+\tan(t)+9}$