Explanation Given: The differential equation is given by, y ″ = 6 t ... ( 1 ) The boundary conditions are given as, y ( 0 ) = 3 , y ′ ( 0 ) = − 1 . Approach: Find the characteristic equation by substituting y ″ = r 2 , y ′ = r . The complementary solution of the differential equation is, y c = c 1 e r 1 t + c 2 e r 2 t + c 3 e r 3 t + c 4 e r 4 t ... ( 2 ) Here, r 1 , r 2 , r 3 and r 4 are the roots of the characteristic equation and c 1 , c 2 , c 3 , and c 4 are the constants. The general solution of the differential equation is given by, y = y c + y p ... ( 3 ) Here, y c is complementary solution and y p is particular solution of the differential equation. Calculation: The auxiliary equation of the equation ( 1 ) is as follows. r 2 = 0 Solve the above equation. r 2 = 0 r = 0 So the repeated roots are 0 , 0 . The complementary solution for the equation ( 1 ) using the roots of auxiliary equation is, y c = c 1 e ( 0 ) t + c 2 t e ( 0 ) t = c 1 + c 2 t Let the particular solution is, y p = t 2 ( A t + B ) = A t 3 + B t 2 ... ( 4 ) Differentiate the above equation with respect to t . y ′ p = d d t ( A t 3 + B t 2 ) = 3 A t 2 + 2 B t Again differentiate the above equation with respect to t . y ″ p = d d t ( 3 A t 2 + 2 B t ) = 6 A t + 2 B Substitute 6 A t + 2 B for y ″ and A t 3 + B t 2 for y in the equation ( 1 )

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 4.5, Problem 24E

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The solution to the initial value problem y″=6t; y(0)=3, y′(0)=−1.

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Chapter 4.5, Problem 23E

Chapter 4.5, Problem 25E

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The solution to the initial value problem y ″ = 6 t ; y ( 0 ) = 3 , y ′ ( 0 ) = − 1 .

Solution Summary: The author explains how to find the solution to the initial value problem by substituting y′′=6tmathrm...(0)

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To find:

The solution to the initial value problem y″=6t; y(0)=3, y′(0)=−1.

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Chapter 4 Solutions

The solution to the initial value problem y ″ = 6 t ; y ( 0 ) = 3 , y ′ ( 0 ) = − 1 .

Solution Summary: The author explains how to find the solution to the initial value problem by substituting y′′=6tmathrm...(0)

Concept explainers

Videos

To find: The solution to the initial value problem y″=6t; y(0)=3, y′(0)=−1.

Want to see the full answer?

Chapter 4 Solutions

To find:

The solution to the initial value problem y″=6t; y(0)=3, y′(0)=−1.