Here is a solution by the method of characteristics. So, we assume that the equation is
$$
u_t+u_x=\cos^2 u,\quad u(x,0)=u_0(x).
$$
Consider the characteristics defined by
$$
\frac{dx}{ds}=1,\quad x(0)=\tau,\\
\frac{dt}{ds}=1,\quad t(0)=0,\\
\frac{du}{ds}=\cos^2 u,\quad u(0)=u_0(\tau).
$$
Obviously, from first two equations
$$
x=s+\tau,\quad t=s\implies \tau=x-t.
$$
From the third equation
$$
\tan u=s+\tan u_0(\tau),
$$
or, finally,
$$
u=\tan^{-1}(\tan u_0(x-t)+t)
$$
as required.
Your teacher is just applying the fundamental theorem of calculus (FTC): if $F'(x) = f(x)$ for all $x \in [a, b]$, then $$\int_a^b f(x)\ dx = F(b) - F(a)$$
It is common when solving first-order ODEs to state things in terms of antiderivatives instead of integrals, but it is not necessary. They are essentially the same thing. And note that the FTC only requires one anti-derivative of $f$, not the entire family. So as long as you know a function $F$ such that $F' = f$, it does not matter that there are others.
In this case, your teacher expressed
$$\frac{y'(t)}{y(t)} = 0,85$$
So,
$$\int_0^t \frac{y'(t)}{y(t)}\ dt = \int_0^t 0,85\ dt = 0,85t$$
Now by the chain rule for integration,
$$\int_0^t \frac{y'(t)}{y(t)}\ dt = \int_{y(0)}^{y(t)} \frac{dy}{y}= \int_{19}^{y(t)} \frac{dy}{y}$$
and by the FTC, since $\frac{d}{dx}(\log |x|) = \frac{1}{x}$, $$\int_{19}^{y(t)} \frac{dy}{y} = \log |y|\left|_{19}^{y(t)}\right. = \log|y(t)| - \log 19$$
$$ \log|y(t)| - \log 19 = 0,85t$$$$\log|y(t)| = 0,85t + \log 19$$
In this case, it is not really true that we have not had to find $C$. We have found it. It is $\log 19$. There is no need to estimate a decimal expression for it, as we aren't planning to use it in this form. We want an equation solved for $y$, so:
$$e^{\log |y(t)|} = e^{0,85t + \log 19} = e^{0,85t}e^{\log 19}$$
$$|y(t)| = 19e^{0,85t}$$
We can go one step further: since $y$ is differentiable, it is also continuous, and as $|y|$ is never zero, $y$ is either always positive, or always negative. Since $y(0) = 19 > 0$, $y$ is always positive. Therefore,
$$y(t) = 19e^{0,85t}$$
Best Answer
$$ \frac{dy}{dx} = \frac{2x}{1+2y}$$
$$ (1+2y)dy = 2x \ dx$$
$$ y^2 + y = x^2 +c$$
To find an explicit solution, you need to factorise the following quadratic to obtain $y$:
$$y^2 + y -x^2-c = 0$$
Use the quadratic formula:
$$y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$a = 1, \ b = 1, \ c= -(x^2+c)$
Once you've obtained an expression for $y$, you can then use the boundary condition to find the constant.