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# lecture_8 (dragged) 1 - 2 MA 36600 LECTURE NOTES FRIDAY...

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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 di ff erential equation becomes 1 y (1 y ) dy dt = 1 y + 1 1 y dy dt = d dt ln | y | ln | 1 y | . Upon integrating both sides of the di ff erential equation, we find that ln | y | ln | 1 y | = r t + C for some constant C . This is a perfectly valid implicit solution to the di ff erential equation, but we want to solve for y = y ( t ) to find an explicit solution. To this end, recall the following identity for the di ff erence of two logarithms: ln A ln B = ln A B . This means we have ln y 1 y = r t + C = y 1 y = C 1 e rt where C 1 = ± e C is a constant. For 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 find the solution y ( t ) = y 0 y 0 + (1 y 0 ) e rt = P ( t ) = K y ( t ) = P 0 K P 0 + ( K P 0 ) e rt . These are explicit solutions to the initial value problems.
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