Explanation The given differential equation is y ″ + 2 y ′ + y = 2 e − t . Consider the homogeneous equation y ″ + 2 y ′ + y = 0 . Obtain the characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: r 2 + 2 r + 1 = 0 . The roots of the characteristic equation are as follows. r 2 + 2 r + 1 = 0 r 1 , 2 = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 1 ) 2 ( 1 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = − 2 ± 4 − 4 2 r 1 , 2 = − 2 ± 0 2 Simplify further, r 1 , 2 = − 2 2 r 1 , 2 = − 1 r = − 1 Note that, the general equation is y c ( t ) = c 1 e r t + c 2 t e r t where r are the double root of the characteristic equation. Substitute r = − 1 in y c ( t ) = c 1 e r t + c 2 t e r t . y c ( t ) = c 1 e − t + c 2 t e − t Thus, the complementary solution is y c ( t ) = c 1 e − t + c 2 t e − t . Compare y ″ + 2 y ′ + y = 2 e − t with standard form y ″ + b y ′ + c y = g ( t ) . Here, b = 2 , c = 1 and g ( t ) = 2 e − t . Substitute b = 2 , c = 1 in r 1 + r 2 = − b , r 1 r 2 = c . r 1 + r 2 = − 2 r 1 r 2 = 1 Thus, the value of r 1 and r 2 is − 1 and − 1 . Obtain the solution of the equation ( D − r 1 ) u = g ( t ) as follows. Substitute r 1 = − 1 and g ( t ) = 2 e − t in ( D − r 1 ) u = g ( t ) . ( D + 1 ) u = 2 e − t d u d t + u = 2 e − t Note that, the general equation is y ′ + p ( x ) y = q ( x ) , and the integral factor is μ = e ∫ p ( x ) d x , then the general solution is y μ = ∫ q ( x ) μ d x . Integral factor is computed as follows. μ = e ∫ 1 d t ( ∵ d u d t + u = 2 e − t and p ( t ) = 1 ) = e t Multiply the equation d u d t + u = 2 e − t by e t

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Question

Chapter 3.5, Problem 38P

To determine

The general solution of the given differential equation.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 3.5, Problem 37P

Chapter 3.5, Problem 39P

Students have asked these similar questions

1 2 21. For the matrix A = 3 4 find AT (the transpose of A). 22. Determine whether the vector @ 1 3 2 is perpendicular to -6 3 2 23. If v1 = (2) 3 and v2 = compute V1 V2 (dot product). .

7. Find the eigenvalues of the matrix (69) 8. Determine whether the vector (£) 23 is in the span of the vectors -0-0 and 2 2

1. Solve for x: 2. Simplify: 2x+5=15. (x+3)² − (x − 2)². - b 3. If a = 3 and 6 = 4, find (a + b)² − (a² + b²). 4. Solve for x in 3x² - 12 = 0. -

Chapter 3 Solutions

Elementary Differential Equations

Knowledge Booster

$Advanced Math$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.