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## diff. equations

I am now doing the same problem, and your thread helped me immensely. I know how to do this type of integration, but I didn't understand what the question was asking. My question is now this, why do I have to put C back into the original equation, and which one, the one I started the problem with, or the one I used for the quadratic formula?

- Nov 4th 2009, 10:47 PM #12
With "C" in the answer, the solution is a general one, that is, the differential equation will be true for any constant C. However, we are looking for a particular solution to the equation, namely, the solution which satisfies P(1) = 2. For this to work, C has to be a certain value. And so we solve for C and plug it into our solution equation (the P(t) equation) to get the particular solution we are looking for.