Question: Can you solve explicitly? please. I don't know how to solve. Thank you for help. 

![enter image description here](https://i.stack.imgur.com/2NiJi.jpg)

Answer: 
I think this is all unnecessarily complicated. You can use induction over the number of edges, without removing edgeless nodes.

Basis step: A forest without edges is bipartite.

Inductive step: Assume all forests with $k$ edges are bipartite. Given a forest $F$ with $k+1$ edges, remove an edge $e$ and find a bipartition of the resulting forest $F'$ with $k$ edges. The removed edge $e$ of $F$ connects two trees in $F'$. If the vertices it connects are in the same class, flip the class of all nodes in one of the two trees. Since there are no edges between the nodes of the tree and other nodes, this again results in a valid bipartition. Now the two nodes of $e$ are in different classes, so adding $e$ back in doesn't invalidate the bipartition.
This seems to be an attempt at a "two dimensional" induction. This means that we no longer have a single base case.

To be consistent with your method, I think it'd be best to choose and arbitrary number of vertices $n \geq 0$, then induct on the number of edges. The base case is the $n$-vertex null graph, and the inductive step is adding edges between two of these $n$ vertices.

From any proper $2$-colouring of a graph, you can obtain another by toggling the colours in any connected component.

To be able to add an edge $ij$ an obtain a forest, the vertices $i$ and $j$ must have already been in distinct connected component (otherwise the introduction of $ij$ adds a cycle). So, if it turns out that $i$ and $j$ have the same colour, we toggle the colours of the connected component containing $j$, then add the edge $ij$.