Explanation The differential equation can be written as: ( 2 − t ) y ″ ′ + ( 2 t − 3 ) y ″ − t y ′ + y = 0 y ″ ′ + ( 2 t − 3 ) ( 2 − t ) y ″ − t ( 2 − t ) y ′ + 1 ( 2 − t ) y = 0 By comparing the differential equation with general form of differential equation ( y ″ ) ′ + p 1 ( t ) y ″ + p 2 ( t ) y ′ + p 3 ( t ) y = 0 , the value of p 1 ( t ) = ( 2 t − 3 ) ( 2 − t ) , p 2 ( t ) = t ( 2 − t ) , and p 3 ( t ) = 1 ( 2 − t ) . Substitute, y = e t v ( t ) in the equation y ″ ′ + ( 2 t − 3 ) ( 2 − t ) y ″ − t ( 2 − t ) y ′ + 1 ( 2 − t ) y = 0 . ( e t v ) ″ ′ + ( 2 t − 3 ) ( 2 − t ) ( e t v ) ″ − t ( 2 − t ) ( e t v ) ′ + 1 ( 2 − t ) ( e t v ) = 0 e t v ″ ′ + ( 3 × e t + ( 2 t − 3 ) ( 2 − t ) ) ( v ) ″ + ( 3 × e t + 2 ( 2 t − 3 ) ( 2 − t ) × e t − t ( 2 − t ) × e t ) ( v ) ′ = 0 e t [ v ″ ′ + ( 3 + ( 2 t − 3 ) ( 2 − t ) ) ( v ) ″ + ( 3 + 2 ( 2 t − 3 ) ( 2 − t ) − t ( 2 − t ) ) ( v ) ′ ] = 0 v ″ ′ + ( 3 + ( 2 t − 3 ) ( 2 − t ) ) ( v ) ″ + ( 3 + 2 ( 2 t − 3 ) ( 2 − t ) − t ( 2 − t ) ) ( v ) ′ = 0 On the further simplification, the following is obtained

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ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 4.1, Problem 27P

To determine

To solve: The differential equation (2−t)y″′+(2t−3)y″−ty′+y=0 with initial condition y1(t)=et.

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Chapter 4.1, Problem 26P

Chapter 4.1, Problem 28P

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Chapter 4 Solutions

Elementary Differential Equations

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The differential equation ( 2 − t ) y ″ ′ + ( 2 t − 3 ) y ″ − t y ′ + y = 0 with initial condition y 1 ( t ) = e t .

To solve: The differential equation (2−t)y″′+(2t−3)y″−ty′+y=0 with initial condition y1(t)=et.

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Chapter 4 Solutions