Explanation Given information: The differential equation is y ' = 2 ( y − 10 ) and initial conditions are y ( 0 ) = 25 and y ( 0 ) = 5 Formula used : ∫ d x ( x ± a ) = log ( x ± a ) + C Calculation: Consider the given differential equation. y ' = 2 ( y − 10 ) Separate the variables as shown below. ⇒ d y d t = 2 ( y − 10 ) ⇒ d y ( y − 10 ) = 2 d t Integrating both sides, we get ∫ d y ( y − 10 ) = 2 ∫ d t ⇒ log ( y − 10 ) = 2 t + k ⇒ y − 10 = e 2 t + k ⇒ y = e k ⋅ e 2 t + 10 ⇒ y = C e 2 t + 10 Using y ( 0 ) = 25 in y = C e

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 10.2, Problem 1E

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To find:The general solution of y'=2(y−10) and two solutions satisfying y(0)=25 and y(0)=5 and to sketch their graph.

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Chapter 10.2, Problem 4PQ

Chapter 10.2, Problem 2E

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The general solution of y ' = 2 ( y − 10 ) and two solutions satisfying y ( 0 ) = 25 and y ( 0 ) = 5 and to sketch their graph.

Solution Summary: The author explains how to find the general solution of y'=2(y-10) and to sketch their graph.

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To find:The general solution of y'=2(y−10) and two solutions satisfying y(0)=25 and y(0)=5 and to sketch their graph.

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Chapter 10 Solutions