115_pdfsam_math 54 differential equation solutions odd

115_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 3.2 16. By definition, p(t + h) − p(t) . h→0 h p (t) = lim Replacing h by −h in the above equation, we obtain p (t) = lim p(t − h) − p(t) p(t) − p(t − h) = lim . h→0 h→0 −h h Adding the previous two equations together yields 2p (t) = lim p(t + h) − p(t) p(t) − p(t − h) + h→0 h h p(t + h) − p(t − h) = lim . h→0 h p(t + h) − p(t − h) . 2h Thus p (t) = lim h→0 19. This problem can be regarded as a compartmental analysis problem for the population of fish. If we let m(t) denote the mass in million tons of a certain species of fish, then the mathematical model for this process is given by dm = increase rate − decrease rate. dt The increase rate of fish is given by 2m million tons/yr. The decrease rate of fish is given as 15 million tons/yr. Substituting these rates into the above equation we obtain dm = 2m − 15, dt m(0) = 7 (million tons). This equation is linear and separable. Using the initial condition, m(0) = 7 to evaluate the arbitrary constant we obtain 1 15 m(t) = − e2t + . 2 2 Knowing this equation we can now find when all the fish will be gone. To determine when all the fish will be gone we set m(t) = 0 and solve for t. This gives 1 15 0 = − e2t + 2 2 111 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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