We know from Chapter 2 that the family of solutions of the differential equation `dP text[/] dt=0.25P` can be described by `Ptext[(]t text[)]=C e^(0.25*t)`, where `C` can be any real number. (Compare this with Activity 2.) To find the particular solution such that `Ptext[(]0 text[)]=2`, we set `2=Ptext[(]0 text[)]=C e^(0.25*0)=C*1=C`. We find that `C=2`, so the specific function `P_1text[(]t text[)]=2e^(0.25 t)` is the particular solution we seek.
Similarly, to find a solution of the form `Ptext[(]t text[)]=C e^(0.25*t)` such that `Ptext[(]1 text[)]=2`, we set `2=Ptext[(]1 text[)]=Ce^0.25`, and we find that `C=2 e^(-0.25)~~1.55760`. It follows that our second solution is (at least approximately) `P_2text[(]t text[)]=1.55760e^(0.25 t)`. These two solutions are plotted in Figure A2.