Don't understand high pass filters?
The key point is that it is 0 in steady state. That is, as \$t\rightarrow\infty\$, \$V_{out}\rightarrow0\$. When you take the differential equation, you're looking at output voltage with respect to time.

Let's try taking a look at an example of this system.
I'm going to simplify \$K=1\$ and \$RC=1\$. Then, we are left with:
$$
\frac{V_{out}}{V_{in}}\left(j\omega\right)=T\left(j\omega\right)=\frac{1}{1-j/\omega}
$$
To simplify our calculations, let's change the form of it and replace \$j\omega\$ with \$s\$. Then, we get:
$$
T\left(s\right)=\frac{s}{s+1}
$$
Taking the step response of this (that is, \$V_{out}\$ when the input \$V_{in}=u\left(t\right)\$ where \$u\left(t\right)\$ is the Heaviside step function) we get this:


Note that this is in time domain, and in steady state, our output voltage is zero. The bode plot comes out as you'd expect (first order high-pass filter):