Explanation Given information: The differential equation is y ' = − 3 ( y − 12 ) and initial conditions are y ( 0 ) = 20 and y ( 0 ) = 0 . Formula used : ∫ d x ( x ± a ) = log ( x ± a ) + C Calculation: Let us consider the given differential equation. y ' = − 3 ( y − 12 ) Separate the variables as shown below. ⇒ d y d t = − 3 ( y − 12 ) ⇒ d y ( y − 12 ) = − 3 d t Integrating both sides, we get ∫ d y ( y − 12 ) = − 3 ∫ d t ⇒ log ( y − 12 ) = − 3 t + k ⇒ y − 12 = e − 3 t + k ⇒ y = e k ⋅ e − 3 t + 12 ⇒ y = C e − 3 t + 12 Hence, y = 12 + C e − 3 t satisfies y ' = − 3 ( y − 12 ) Using

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 10.2, Problem 2E

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To verify: y=12+Ce−3t satisfies y'=−3(y−12) for all Cand to find two solutions satisfying y(0)=20 and y(0)=0 and to sketch their graph.

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Chapter 10.2, Problem 1E

Chapter 10.2, Problem 3E

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To verify : y = 12 + C e − 3 t satisfies y ' = − 3 ( y − 12 ) for all Cand to find two solutions satisfying y ( 0 ) = 20 and y ( 0 ) = 0 and to sketch their graph.

Solution Summary: The author analyzes the differential equation y'=-3(y-12) and its initial conditions.

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To verify: y=12+Ce−3t satisfies y'=−3(y−12) for all Cand to find two solutions satisfying y(0)=20 and y(0)=0 and to sketch their graph.

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Chapter 10 Solutions