Would you like to explain whether this equation can be solved and how?
In[1308]:= eq1 =
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T] == 1 - C2[R1, T] +
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]/R1 + (Rv
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T])/R1 +
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]
Out[1308]=
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"0", ",", "1"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T] == 1 - C2[R1, T] +
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]/R1 + (Rv
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T])/R1 +
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"2", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R1, T]
In[1309]:= eq2 = Rv C2[1, T] +
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[1, T] == 0
Out[1309]= Rv C2[1, T] +
\!\(\*SuperscriptBox[\(C2\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[1, T] == 0