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2 MA 36600 LECTURE NOTES: FRIDAY, JANUARY 30 We will choose these constants so that A = 1 and B A = 0 i.e., we choose A = B = 1. Simply put, the left-hand side of the diFerential equation becomes 1 y (1 y ) dy dt = ° 1 y + 1 1 y ± dy dt = d dt ² ln | y |− ln | 1 y | ³ . Upon integrating both sides of the diFerential equation, we ±nd that ln | y |− ln | 1 y | = rt + C for some constant C . This is a perfectly valid implicit solution to the diFerential equation, but we want to solve for y = y ( t ) to ±nd an explicit solution. To this end, recall the following identity for the diFerence of two logarithms: ln A ln B =ln A B . This means we have ln ´ ´ ´ ´ y 1 y ´ ´ ´ ´ = rt + C = y 1 y = C 1 e rt where C 1 = ± e C is a constant. ²or the initial condition, set t = 0: C 1 = y 0 1 y 0 = y ( t ) 1 y ( t ) = y 0 1 y 0 e rt Solving for y = y ( t ), we ±nd the solution y ( t )= y 0 y 0 +(1 y 0 ) e rt = P ( t )= Ky ( t )= P 0 K P 0 +( K P 0 ) e rt .
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