Answer
$$
\frac{d u}{d t}=\frac{2 t+\sec ^{2} t}{2 u}, \quad u(0)=-5 .
$$
the solution of the given differential equation is
$$
u=- \sqrt {t^{2}+\tan t+25}.
$$
Work Step by Step
$$
\frac{d u}{d t}=\frac{2 t+\sec ^{2} t}{2 u}, \quad u(0)=-5 .
$$
We write the equation in terms of differentials:
$$
2 u d u=\left(2 t+\sec ^{2} t\right) d t
$$
and integrate both sides:
$$
\int 2 u d u=\int\left(2 t+\sec ^{2} t\right) d t
$$
$ \Rightarrow$
$$
u^{2}=t^{2}+\tan t+C
$$
where C is an arbitrary constant.
To satisfy the initial condition $ u(0)=-5, $ we must have
$$
(-5)^{2}=(0)^{2}+\tan (0)+C \Rightarrow c=25
$$
Therefore,
$$
u^{2}=t^{2}+\tan t+25,
$$
so,
$$
u=\pm \sqrt {t^{2}+\tan t+25},
$$
since, $ u(0)=-5 $, we can write the solution in the form:
$$
u=- \sqrt {t^{2}+\tan t+25}.
$$
