Explanation Given: The differential equation is d p d t + 2 t p = p + 4 t − 2 . Procedure used: Linear equation in standard form: d p d t + P ( t ) p = Q ( t ) The integrating factor of the linear equation: I F = e ∫ P ( t ) d t Solution of linear differential equation: p ⋅ I F = ∫ Q ( t ) ⋅ I F d t Calculation: Consider the given differential equation, d p d t + 2 t p = p + 4 t − 2 Rearrange the given differential equation. d p d t + 2 t p − p = 4 t − 2 d p d t + ( 2 t − 1 ) p = 4 t − 2 Compare the equation with the standard form d p d t + P ( t ) p = Q ( t ) . P ( t ) = 2 t − 1 and Q ( t ) = 4 t − 2 . Write the expression for integrating factor. I F = e ∫ P ( t ) d t (1) Substitute 2 t − 1 for P ( t ) in the equation (1). I F = e ∫ ( 2 t − 1 ) d t = e ( t 2 − t ) Write the expression for solution of p

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Differential Equations

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Chapter 2.3, Problem 22E

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The general solution of given differential equation, the interval over which the solution is defined and transient terms in general solution.

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Chapter 2.3, Problem 21E

Chapter 2.3, Problem 23E

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The general solution of given differential equation, the interval over which the solution is defined and transient terms in general solution.

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The general solution of given differential equation, the interval over which the solution is defined and transient terms in general solution.

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Chapter 2 Solutions