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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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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Ch. 2.1 - Prob. 11P Ch. 2.1 - Prob. 12P Ch. 2.1 - Prob. 13P Ch. 2.1 - Prob. 14P Ch. 2.1 - Prob. 15P Ch. 2.1 - Prob. 16P Ch. 2.1 - Prob. 17P Ch. 2.1 - Prob. 18P Ch. 2.1 - Prob. 19P Ch. 2.1 - Prob. 20P Ch. 2.1 - Prob. 21P Ch. 2.1 - Prob. 22P Ch. 2.1 - Prob. 23P Ch. 2.1 - Prob. 24P Ch. 2.1 - Prob. 25P Ch. 2.1 - Prob. 26P Ch. 2.1 - Prob. 27P Ch. 2.1 - Variation of Parameters. Consider the following...Ch. 2.1 - Prob. 29P Ch. 2.1 - In each of Problems 29 and 30, use the method of...Ch. 2.2 - Prob. 1P Ch. 2.2 - Prob. 2P Ch. 2.2 - Prob. 3P Ch. 2.2 - Prob. 4P Ch. 2.2 - Prob. 5P Ch. 2.2 - Prob. 6P Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - In each of Problems 1 through 8, solve the given...Ch. 2.2 - Prob. 9P Ch. 2.2 - Prob. 10P Ch. 2.2 - Prob. 11P Ch. 2.2 - Prob. 12P Ch. 2.2 - Prob. 13P Ch. 2.2 - Prob. 14P Ch. 2.2 - Prob. 15P Ch. 2.2 - Prob. 16P Ch. 2.2 - Prob. 17P Ch. 2.2 - Prob. 18P Ch. 2.2 - Prob. 19P Ch. 2.2 - Prob. 20P Ch. 2.2 - Prob. 21P Ch. 2.2 - Prob. 22P Ch. 2.2 - Prob. 23P Ch. 2.2 - Prob. 24P Ch. 2.2 - Prob. 25P Ch. 2.2 - Prob. 26P Ch. 2.2 - Prob. 27P Ch. 2.2 - Prob. 28P Ch. 2.2 - Prob. 29P Ch. 2.2 - Prob. 30P Ch. 2.2 - Prob. 31P Ch. 2.3 - Prob. 1P Ch. 2.3 - Prob. 2P Ch. 2.3 - Prob. 3P Ch. 2.3 - Prob. 4P Ch. 2.3 - Prob. 5P Ch. 2.3 - Prob. 6P Ch. 2.3 - Prob. 7P Ch. 2.3 - Prob. 8P Ch. 2.3 - Prob. 9P Ch. 2.3 - Prob. 10P Ch. 2.3 - Prob. 11P Ch. 2.3 - Prob. 12P Ch. 2.3 - Prob. 13P Ch. 2.3 - Prob. 14P Ch. 2.3 - Prob. 15P Ch. 2.3 - Prob. 16P Ch. 2.3 - Assume that the conditions are as in Problem 16...Ch. 2.3 - Prob. 18P Ch. 2.3 - Prob. 19P Ch. 2.3 - Prob. 20P Ch. 2.3 - Prob. 21P Ch. 2.3 - Prob. 22P Ch. 2.3 - Prob. 23P Ch. 2.3 - Prob. 24P Ch. 2.4 - In each of Problems 1 through 6, determine...Ch. 2.4 - Prob. 2P Ch. 2.4 - Prob. 3P Ch. 2.4 - Prob. 4P Ch. 2.4 - Prob. 5P Ch. 2.4 - Prob. 6P Ch. 2.4 - Prob. 7P Ch. 2.4 - Prob. 8P Ch. 2.4 - Prob. 9P Ch. 2.4 - Prob. 10P Ch. 2.4 - Prob. 11P Ch. 2.4 - Prob. 12P Ch. 2.4 - Prob. 13P Ch. 2.4 - Prob. 14P Ch. 2.4 - Prob. 15P Ch. 2.4 - Prob. 16P Ch. 2.4 - Prob. 17P Ch. 2.4 - Prob. 18P Ch. 2.4 - Prob. 19P Ch. 2.4 - Prob. 20P Ch. 2.4 - Prob. 21P Ch. 2.4 - Prob. 22P Ch. 2.4 - Prob. 23P Ch. 2.4 - Prob. 24P Ch. 2.4 - Prob. 25P Ch. 2.4 - Prob. 26P Ch. 2.4 - Prob. 27P Ch. 2.5 - Prob. 1P Ch. 2.5 - Prob. 2P Ch. 2.5 - Prob. 3P Ch. 2.5 - Prob. 4P Ch. 2.5 - Prob. 5P Ch. 2.5 - Prob. 6P Ch. 2.5 - Prob. 7P Ch. 2.5 - Prob. 8P Ch. 2.5 - Prob. 9P Ch. 2.5 - Prob. 10P Ch. 2.5 - Prob. 11P Ch. 2.5 - Prob. 12P Ch. 2.5 - Prob. 13P Ch. 2.5 - Prob. 14P Ch. 2.5 - Prob. 15P Ch. 2.5 - Prob. 16P Ch. 2.5 - Prob. 17P Ch. 2.5 - Prob. 18P Ch. 2.5 - Prob. 19P Ch. 2.5 - Prob. 20P Ch. 2.5 - Prob. 21P Ch. 2.5 - Prob. 22P Ch. 2.5 - Prob. 23P Ch. 2.5 - Prob. 24P Ch. 2.5 - Prob. 25P Ch. 2.5 - Prob. 26P Ch. 2.5 - Prob. 27P Ch. 2.6 - Prob. 1P Ch. 2.6 - Prob. 2P Ch. 2.6 - Prob. 3P Ch. 2.6 - Prob. 4P Ch. 2.6 - Prob. 5P Ch. 2.6 - Prob. 6P Ch. 2.6 - Prob. 7P Ch. 2.6 - Prob. 8P Ch. 2.6 - Prob. 9P Ch. 2.6 - Prob. 10P Ch. 2.6 - Prob. 11P Ch. 2.6 - Prob. 12P Ch. 2.6 - Prob. 13P Ch. 2.6 - Prob. 14P Ch. 2.6 - Prob. 15P Ch. 2.6 - Prob. 16P Ch. 2.6 - Prob. 17P Ch. 2.6 - Prob. 18P Ch. 2.6 - Prob. 19P Ch. 2.6 - Prob. 20P Ch. 2.6 - Prob. 21P Ch. 2.6 - Prob. 22P Ch. 2.7 - Prob. 1P Ch. 2.7 - Prob. 2P Ch. 2.7 - Prob. 3P Ch. 2.7 - Prob. 4P Ch. 2.7 - Prob. 5P Ch. 2.7 - Prob. 6P Ch. 2.7 - Prob. 7P Ch. 2.7 - Prob. 8P Ch. 2.7 - Prob. 9P Ch. 2.7 - Prob. 10P Ch. 2.7 - Prob. 11P Ch. 2.7 - Prob. 12P Ch. 2.7 - Prob. 14P Ch. 2.7 - Prob. 15P Ch. 2.7 - Prob. 16P Ch. 2.7 - Prob. 17P Ch. 2.8 - Prob. 1P Ch. 2.8 - Prob. 2P Ch. 2.8 - Prob. 3P Ch. 2.8 - Prob. 4P Ch. 2.8 - Prob. 5P Ch. 2.8 - Prob. 6P Ch. 2.8 - Prob. 7P Ch. 2.8 - Prob. 8P Ch. 2.8 - Prob. 9P Ch. 2.8 - Prob. 10P Ch. 2.8 - Prob. 11P Ch. 2.8 - Prob. 12P Ch. 2.8 - Prob. 13P Ch. 2.8 - Prob. 14P Ch. 2.8 - Prob. 15P Ch. 2.8 - Prob. 16P Ch. 2.8 - Prob. 17P Ch. 2.8 - Prob. 18P Ch. 2.9 - Prob. 1P Ch. 2.9 - Prob. 2P Ch. 2.9 - Prob. 3P Ch. 2.9 - Prob. 4P Ch. 2.9 - Prob. 5P Ch. 2.9 - Prob. 6P Ch. 2.9 - Prob. 7P Ch. 2.9 - Prob. 8P Ch. 2.9 - Prob. 9P Ch. 2.9 - Prob. 10P Ch. 2 - Prob. 1MP Ch. 2 - Prob. 2MP Ch. 2 - Prob. 3MP Ch. 2 - Prob. 4MP Ch. 2 - Prob. 5MP Ch. 2 - Prob. 6MP Ch. 2 - Prob. 7MP Ch. 2 - Prob. 8MP Ch. 2 - Prob. 9MP Ch. 2 - Prob. 10MP Ch. 2 - Prob. 11MP Ch. 2 - Prob. 12MP Ch. 2 - Prob. 13MP Ch. 2 - Prob. 14MP Ch. 2 - Prob. 15MP Ch. 2 - Prob. 16MP Ch. 2 - Prob. 17MP Ch. 2 - Prob. 18MP Ch. 2 - Prob. 19MP Ch. 2 - Prob. 20MP Ch. 2 - Prob. 21MP Ch. 2 - Prob. 22MP Ch. 2 - Prob. 23MP Ch. 2 - Prob. 24MP Ch. 2 - Prob. 25MP Ch. 2 - Prob. 28MP Ch. 2 - Prob. 29MP Ch. 2 - Prob. 31MP Ch. 2 - Prob. 32MP Ch. 2 - Prob. 33MP Ch. 2 - Prob. 34MP Ch. 2 - Prob. 35MP Ch. 2 - Prob. 36MP Ch. 2 - Prob. 37MP

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.

Want to see the full answer?

Chapter 2 Solutions