In the figure, the controlled sources of 8Ω, 4Ω and 4U2 are all connected in series, and the branch injects current into the node. The current is (4U2-U2)/(8+4) = (4U2-U2)/12, and the current flows into the node.

6 Ω series 15V voltage source current is (U2- 15)/6, and the direction is the outflow node;

12 Ω resistance branch, that is, the current in the branch where U2 is located is U2/ 12, and the direction is the inflow node.

According to KCl, (U2-15)/6+U2/12 = (4u2-U2)/12.

The result is: (1/6+112+1/2) U2 =15/6+4u2/12, which is the figure.

If you don't understand it well, you can also equate the 15V voltage source and 4U2 controlled voltage source as current sources according to the power equivalent transformation, that is, 15/6A and 4U2/(8+4)=4U2/ 12(A), which are the current injected into the node; Other resistance branches are the current flowing from the node, namely U2/6, U2/ 12 and U2/ 12.

But this equation is useless, because through this equation, you also find that U2 can't be calculated, that is, U2 doesn't exist, so you can't find Uoc, which only shows that there is no Thevenin equivalent circuit in the original circuit, only Norton circuit.