Explanation Given: e x ⋅ y ′ = 1 + 2 e x ⋅ y Concept used: The general solution of the first order linear differential equation of the form y ′ + P y = Q , where P and Q are constants or functions of x only is, y ⋅ ( IF ) = ∫ Q ⋅ ( IF ) d x + c where IF = integrating factor = e ∫ P d x and c is any arbitrary constant . Calculation: Given differential equation is e x ⋅ y ′ = 1 + 2 e x ⋅ y Dividing both sides of the equation by e x , we get e x ⋅ y ′ e x = 1 + 2 e x ⋅ y e x ⇒ y ′ = 1 e x + 2 y ⇒ y ′ − 2 y = e − x Now comparing the linear differential equation y ′ − 2 y = e − x with y ′ + P y = Q , we get P = − 2 and Q = e − x

ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

4th Edition

ISBN: 9781319411657

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 9.5, Problem 17E

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To find the general solution of the linear differential equation ex⋅y′=1+2ex⋅y.

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Chapter 9.5, Problem 16E

Chapter 9.5, Problem 18E

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Chapter 9 Solutions

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To find the general solution of the linear differential equation e x ⋅ y ′ = 1 + 2 e x ⋅ y .

Solution Summary: The author explains the general solution of the first order linear differential equation excdot

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To find the general solution of the linear differential equation ex⋅y′=1+2ex⋅y.

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Chapter 9 Solutions