Explanation The given differential equation is y ′ ( t ) = y ( y − 2 ) , for t ≥ 0 . That is, d y d t = y ( y − 2 ) . Separate the y and t variables. d y y ( y − 2 ) = d t Integrate both sides of the equation. ∫ d y y ( y − 2 ) = ∫ d t ∫ 1 2 ( 1 ( y − 2 ) − 1 y ) d y = t + C 1 2 ( ln | y − 2 | − ln | y | ) = t + C 1 ln ( y − 2 y ) = 2 t + 2 C 1 Take exponential on both sides of the above expression. That is, y − 2 y = e ( 2 t + C ) . Determine the value of arbitrary constant C using initial condition y ( 0 ) = 2.01 . 2.01 − 2 2.01 = e ( 2 ( 0 ) + C ) 0.004975 = e ( C ) ln ( 0.004975 ) = C C = − 5.303 Therefore, the required solution of the differential equation y ′ ( t ) = y ( y − 2 ) is y − 2 y = e 2 t − 5

Calculus: Early Transcendentals and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition) (Briggs, Cochran, Gillett & Schulz, Calculus Series)

3rd Edition

ISBN: 9780134995991

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 9.2, Problem 2QC

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To find: Whether the solution to the equation y′(t)=y(y−2) with the initial conditions y(0)=2.01 and y(1)=−1 is increasing or decreasing.

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Calculus: Early Transcendentals and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition) (Briggs, Cochran, Gillett & Schulz, Calculus Series)

Ch. 9.1 - What are the orders of the equations in Example 2?...Ch. 9.1 - What is the solution of the initial value problem...Ch. 9.1 - Solve the initial value problem in Example 4a with...Ch. 9.1 - Suppose the initial conditions in Example 5a are...Ch. 9.1 - In Example 7, if the height function were given by...Ch. 9.1 - Consider the differential equation y(t) + 9y(t) =...Ch. 9.1 - If the general solution of a differential equation...Ch. 9.1 - Does the function y(t) = 2t satisfy the...Ch. 9.1 - Does the function y(t) = 6e3t satisfy the initial...Ch. 9.1 - The solution to the initial value problem y(t) = 2...

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Whether the solution to the equation y ′ ( t ) = y ( y − 2 ) with the initial conditions y ( 0 ) = 2.01 and y ( 1 ) = − 1 is increasing or decreasing.

Solution Summary: The author explains how to determine the value of arbitrary constant C using the initial condition y(0)=2.01.