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Unformatted text preview: learning decreases.) Thus dP dt = k ( MP ( t )) where k is a positive constant. Solve this linear diﬀerential equation. We ﬁrst rewrite the equation as dP dt + kP ( t ) = kM, which is a linear equation. The integration factor is I ( t ) = exp( Z k dt ) = e kt . So the solution is given by P ( t ) = 1 I ( t ) Z kMI ( t ) dt = ekt ( Me kt + C ) = M + Cekt for some constant C ....
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This note was uploaded on 05/09/2008 for the course MATH 2130 taught by Professor Dorais during the Spring '08 term at Cornell.
 Spring '08
 DORAIS
 Calculus, Derivative

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