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Chapter 6: Q. 75 (page 572)

Prove Theorem 6.20 by solving the initial-value problem dQdt=kQwithQ(0)=Q0, where k is a constant

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Step by step solution

01

Step 1. Given

The given problem isdQdt=kQwithQ(0)=Q0.

02

Step 2. Proof

鈭�dQQ=鈭�kdtlnQ=kt+CQ=ekt+C=AektNowusetheinitialconditionQ=Q0fort=0intheaboveexpressiontoobtainthesolutionoftheinitial-valueproblemasQ0=ASubstitutethisvalueoftheconstantAinthesolutionofthedifferentialequationtoobtainthesolutionoftheinitial-valueproblemasQ(t)=Q0ektObserve that the differential equation does not contain the independent variable at all. So, solve the differential equation by using antidifferentiation method

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