{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

200_pdfsam_math 54 differential equation solutions odd

# 200_pdfsam_math 54 differential equation solutions odd - y...

This preview shows page 1. Sign up to view the full content.

Chapter 4 35. We look for a particular solution of the form y p ( t ) = t s ( A 1 t + A 0 ) e t , and choose s = 1 because the auxiliary equation, r 3 + r 2 2 = ( r 1)( r 2 + 2 r + 2) = 0 has r = 1 as a simple root. Hence, y p ( t ) = t ( A 1 t + A 0 ) e t = ( A 1 t 2 + A 0 t ) e t y p ( t ) = A 1 t 2 + (2 A 1 + A 0 ) t + A 0 e t y p ( t ) = A 1 t 2 + (4 A 1 + A 0 ) t + (2 A 1 + 2 A 0 ) e t y p ( t ) = A 1 t 2 + (6 A 1 + A 0 ) t + (6 A 1 + 3 A 0 ) e t y + y 2 y = [10 A 1 t + (8 A 1 + 5 A 0 )] e t = te t . Equating the corresponding coeﬃcients, we find that 10 A 1 = 1 , 8 A 1 + 5 A 0 = 0 A 1 = 1 / 10 , A 0 = 8 A 1 / 5 = 4 / 25 y p ( t ) = 1 10 t 2 4 25 t e t . EXERCISES 4.5: The Superposition Principle and Undetermined Coeﬃcients Revisited, page 192 1. Let g 1 ( t ) := sin t and g 2 ( t ) := e 2 t . Then y 1 ( t ) = cos t is a solution to y y + y = g 1 ( t ) and y 2 ( t ) = e 2 t / 3 is a solution to y y + y = g 2 ( t ) . (a) The right-hand side of the given equation is 5 sin t = 5 g
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: y ( t ) = 5 y 1 ( t ) = 5 cos t is a solution to y − y + y = 5 sin t . (b) We can express sin t − 3 e 2 t = g 1 ( t ) − 3 g 2 ( t ). So, by the superposition principle the desired solution is y ( t ) = y 1 ( t ) − 3 y 2 ( t ) = cos t − e 2 t . (c) Since 4 sin t + 18 e 2 t = 4 g 1 ( t ) + 18 g 2 ( t ), the function y ( t ) = 4 y 1 ( t ) + 18 y 2 ( t ) = 4 cos t + 6 e 2 t is a solution to the given equation. 196...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online