How do I find the UTP and LTP of an Oscillator?
The Thevenin to the (+) input is:

$$\begin{align*}
V_+ &amp;= \frac{V_o\cdot 6.8k\:\Omega\cdot 10k\:\Omega+16\:\textrm{V}\cdot 6.8k\:\Omega\cdot 5.6k\:\Omega}{6.8k\:\Omega\cdot 10k\:\Omega + 6.8k\:\Omega\cdot 5.6k\:\Omega+5.6k\:\Omega\cdot 10k\:\Omega} \\
 \\
&amp;= 0.419545903\cdot V_o + 3.75913129\:\textrm{V}
\end{align*}$$

Assuming your opamp is rail to rail output and is acting as a comparator, then this means that \$V_+\approx +11.3\:\textrm{V}\$ or \$V_+\approx-3.8\:\textrm{V}\$, depending on the value of \$V_o\$. Let's call these two different voltages, \$V_H = +11.3\:\textrm{V}\$ and \$V_L= -3.8\:\textrm{V}\$.

So the capacitor voltage (and \$V_-\$) will be bouncing between these two values. When it is at \$V_L\$ the output will switch to \$+18\:\textrm{V}\$ and the voltage will start to rise. When it is at \$V_H\$ the output will switch to \$-18\:\textrm{V}\$ and the voltage will start to sink.

There will be two such timing equations:

$$\begin{align*}
V_{RISE(t)} &amp;= V_L+\left(18\:\textrm{V} - V_L\right)\cdot\left(1-e^{\frac{-t}{R C}}\right) \\
 \\
V_{FALL(t)} &amp;= V_H+\left(-18\:\textrm{V} - V_H\right)\cdot\left(1-e^{\frac{-t}{R C}}\right) \\
\end{align*}$$

Setting \$V_{RISE(t)}=V_H\$ and \$V_{FALL(t)}=V_L\$, these are easily solved for time as:

$$\begin{align*}
t_{RISE} &amp;= -R C\cdot\textrm{ln}\left(1-\frac{V_H-V_L}{18\:\textrm{V} - V_L}\right) \approx 449\:\mu\textrm{s} \\
 \\
t_{FALL} &amp;= -R C\cdot\textrm{ln}\left(1-\frac{V_L-V_H}{-18\:\textrm{V} - V_H}\right) \approx 276\:\mu\textrm{s}
\end{align*}$$

The frequency is then about \$\tfrac{1}{449\:\mu\textrm{s}+276\:\mu\textrm{s}}\approx 1380\:\textrm{Hz}\$

With an opamp that doesn't support rail to rail outputs, the thresholds will be different and so will the timing. Also, opamps have slew rate limitations and current limitations, too. But your opamp is ideal.