Explanation The given initial value problem is y ′ ( t ) = t 2 1 − t . Take the indefinite integral of both sides of the differential equation with respect to t. ∫ y ′ ( t ) d t = ∫ ( t 2 1 − t ) d t y ( t ) = ∫ ( t 2 1 − t ) d t Obtain the indefinite integral of ∫ ( t 2 1 − t ) d t by substitution method. Let the variable v be 1 − t . Differentiate v with respect to t , d d t ( v ) = d d t ( 1 − t ) d v d t = − 1 d v = − d t Substitute v = 1 − t and d v = − d t in the integral ∫ ( t 2 1 − t ) d t as, ∫ ( t 2 1 − t ) d t = ∫ ( 1 − v ) 2 v ( − d v ) = − ∫ ( 1 − 2 v + v 2 v ) d v = − ∫ ( 1 v − 2 + v ) d v = − ( ln | v | − 2 v + v 2 2 ) + C Now replace v with 1 − t on the above expression as follows, ∫ ( t 2 1 − t ) d t = − ( ln | 1 − t | − 2 ( 1 − t ) + ( 1 − t ) 2 2 ) + C = − ln | 1 − t | + 2 ( 1 − t ) − ( 1 − t ) 2 2 + C = 2 − 2 t − ln | 1 − t | − ( 1 − t ) 2 2 + C That is, y ( t ) = 2 − 2 t − ln | 1 − t | − ( 1 − t ) 2 2 + C

Single Variable Calculus Format: Unbound (saleable)

3rd Edition

ISBN: 9780134765761

Author: Briggs, William L.^cochran, Lyle^gillett, Bernard^

Publisher: Prentice Hall

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Chapter 9.1, Problem 6E

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To explain: The reason for the graph of the solution of an initial value problem y′(t)=t21−t with the initial condition of y(−1)=ln2 not cross the line t=1.

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Chapter 9.1, Problem 5E

Chapter 9.1, Problem 7E

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The reason for the graph of the solution of an initial value problem y ′ ( t ) = t 2 1 − t with the initial condition of y ( − 1 ) = ln 2 not cross the line t = 1 .

Solution Summary: The author explains the graph of the solution of an initial value problem with the initial condition of y(t)=mathrmln2.

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To explain: The reason for the graph of the solution of an initial value problem y′(t)=t21−t with the initial condition of y(−1)=ln2 not cross the line t=1.

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