Differential Equations (Seperation of Variables)
Posted on 2006-05-26
Find and Sketch the solution to satisfying P = 5 when t = 0.
-- = 2P - 2Pt
The first part of this problem makes plenty of sense:
--- = 2 - 2t dt
ln |P| = 2t - t^2 + C
|P| = e^(2t - t^2 + C)
|P| = e^C * e^(2t - t^2)
A = e^C
|P| = Ae^(2t - t^2)
Now this next part in the book doesnt make sense to me:
"so A > 0 (Makes Sense). In addition, A = 0 gives a solution (WHAT?? HOW?) Thus the general solution to the differential equation is P = Be^(2t - t^2) for any B (Again how so?)"
Then they go on to find the value of B by plugging in t = 0 and P = 5. (this seems easy enough).
I dont get where they just threw in the B from or what they are talking about when they say A = 0 is a solution.