Answer
$\displaystyle u = -\sqrt{t^2+\tan(t)+25}$
Work Step by Step
$\displaystyle \frac{du}{dt} = \frac{2t+\sec^2(t)}{2u}$
Separation of variables:
$\displaystyle\int 2u\cdot du = \int [2t + \sec^2(t)]dt $
$\displaystyle u^2 = t^2+\tan(t)+C$
Plug in initial condition to solve for $C$:
$\displaystyle (-5)^2 = (0)^2 + \tan(0) + C$
$\displaystyle25 = 0 + 0+C = C$
Plug in the value for $C$ that was just found and isolate $u$
$\displaystyle u^2 = t^2+\tan(t)+25$
$\displaystyle u =\pm\sqrt{t^2+\tan(t)+25}$
Now that we have two possible solutions with the plus or minus, we must decide which sign is appropriate. To find out, we have to plug the initial condition into each and see what works. Remember, the initial condition is $u(0) = -5$
Positive Solution: $\qquad\qquad\qquad\qquad\qquad\qquad$ Negative Solution:
$\displaystyle u = +\sqrt{t^2+\tan(t)+25}\qquad\qquad\qquad u = -\sqrt{t^2 +\tan(t)+25}$
$\displaystyle -5 = \sqrt{0 + 0 + 25}\qquad\qquad\qquad\qquad-5 = -\sqrt{(0)+0+25}$
$\displaystyle -5 = 5\qquad\qquad\qquad\qquad\qquad\qquad\quad-5 = -5$
We see that the negative solution is the one that fits the initial condition, so we can drop the positive solution. Our final answer is:
$\displaystyle u = -\sqrt{t^2+\tan(t)+25}$
