I typed

TeXForm[heqn1/.{DiracDelta[x_]:>\[Delta][x]}]

to hopefully get back something which matches your equation. $$\begin{multline*} k_{\text{CLS}}\left(T _D^{(0,2,0)}(x,y,t)+T _D^{(2,0,0)}(x,y,t)\right)+\delta (x-\text{x1}) \delta (x-\text{x2}) \delta (y-\text{y1})\delta (y-\text{y2})\\ +\delta (x-\text{x1}) \delta(y-\text{y1})+\delta (x-\text{x2}) \delta(y-\text{y2})=\\ \text{Cp} _{\text{CLS}} \rho _{\text{CLS}}T _D^{(0,0,1)}(x,y,t) \end{multline*}$$ This looks right. If you want to keep these new variable names (y1 instead of b1...), but still be sure of equivalence, you could do more substitutions besides DiracDelta and end up with almost equivalent tex code. Converting the indexed Derivative to Newton notation is possible, but finicky -- I think it reads just fine in this case.