Calculus II

Class Notes, 04/07/99


In the figure below we solve the equation indicated in the first line with the initial condition that Q = 10 when t = 0.

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In the third line below we obtain the solutions the absolute value equation, assuming that C is a positive constant.

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We use similar techniques to solve the equation in the figure below, with the given initial condition.

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Applying the initial condition, we obtain C = -.9e for the R < 1 solution and C = .9e for the R > 1 solution.

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We solve the equation below by separating the variables in the usual manner.

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We consider a 1000 gallon container of water in which is dissolved 10 pounds of salt.

We wish to determine the amount of salt as a function of time. To do so we will set up a differential equation involving the rate dS / dt at which the salt leaves the system.

More generally, if S is the amount of salt the concentration of salt will be S / V, where V is the volume of the container.

The equation dS / dt = .003 S, S(0) = 10 is easily solved by the usual methods, yielding S(t) = 10 e(-.003t).

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