Explanation Given: The differential equation is given by, y ′ − y = 1... ( 1 ) y ( 0 ) = 0 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 − 1 = 0 Solve the above equation. r − 1 = 0 r = 1 The complementary solution for the equation ( 1 ) using the roots of auxiliary equation is, y c = c 1 e t Let the particular solution is, y p = k

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 23E

To determine

To find:

The solution to the initial value problem y′−y=1, y(0)=0.

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Chapter 4.5, Problem 22E

Chapter 4.5, Problem 24E

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Fundamentals of Differential Equations and Boundary Value Problems

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