Explanation The given initial value problem is: ( 1 + t 2 ) y ″ + 2 t y ' + 3 t − 2 = 0 , y ( 1 ) = 2 , y ' ( 1 ) = − 1 From the given differential equation is: ( 1 + t 2 ) y ″ + 2 t y ' + 3 t − 2 = 0 Let y ' = v ( t ) and differentiate the equation d y d t = v with respect to t : y ' = v d d t ( d y d t ) = d d t ( v ) d 2 y d t 2 = d v d t Therefore, the value of second derivative is y ″ = d v d t . Substitute y ' = v and y ″ = d v d t in ( 1 + t 2 ) y ″ + 2 t y ' + 3 t − 2 = 0 . ( 1 + t 2 ) d v d t + 2 t v + 3 t 2 = 0 d v d t + 2 t 1 + t 2 v + 3 t 2 ( 1 + t 2 ) = 0 d v d t + 2 t 1 + t 2 v = − 3 t 2 ( 1 + t 2 ) The above equation shows the first order differential equation. So, calculate the integrating factor of the first order differential equation d v d t + 2 t 1 + t 2 v = − 3 t 2 ( 1 + t 2 ) . μ ( t ) = exp ∫ 2 t 1 + t 2 d t = e ln ( 1 + t 2 ) = 1 + t 2 Multiply equation d v d t + 2 t 1 + t 2 v = − 3 t 2 ( 1 + t 2 ) with μ ( t ) . v μ ( t ) = ∫ μ ( t ) − 3 t 2 ( 1 + t 2 ) d t v ( 1 + t 2 ) = − 3 ∫ ( 1 + t 2 ) t 2 ( 1 + t 2 ) d t = − 3 ∫ 1 t 2 d t = 3 t + c 1 On further simplification, the following is obtained: v t ( 1 + t 2 ) = 3 + t c 1 Form the initial conditions y ' ( 1 ) = − 1 . When t = 1 then y ' = − 1 and when t = 1 then v = − 1 . Substitute the values in v t ( 1 + t 2 ) = 3 + t c 1

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Question

Chapter 2, Problem 37MP

To determine

To solve: The given initial value problem (1+t2)y″+2ty'+3t−2=0, y(1)=2, y'(1)=−1.

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Chapter 2, Problem 36MP

Chapter 3.1, Problem 1P

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Chapter 2 Solutions

Elementary Differential Equations and Boundary Value Problems, Enhanced

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Knowledge Booster

The given initial value problem ( 1 + t 2 ) y ″ + 2 t y ' + 3 t − 2 = 0 , y ( 1 ) = 2 , y ' ( 1 ) = − 1 .

To solve: The given initial value problem (1+t2)y″+2ty'+3t−2=0, y(1)=2, y'(1)=−1.

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Chapter 2 Solutions