{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

253_pdfsam_math 54 differential equation solutions odd

# 253_pdfsam_math 54 differential equation solutions odd -...

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

Review Problems 17. To solve the auxiliary equation, r 3 + 10 r 11 = 0, we note that r 1 = 1 is a root. Dividing the polynomial r 3 + 10 r 11 by r 1 we get r 3 + 10 r 11 = ( r 1)( r 2 + r + 11) , and so the other two roots are r 2 , 3 = 1 ± 1 4(1)(11) 2 = 1 2 ± 43 2 i. A general solution is then given by y ( t ) = c 1 e t + e t/ 2 c 2 cos 43 t 2 + c 3 sin 43 t 2 . 19. By inspection, we find that r = 3 as a root of the auxiliary equation, 4 r 3 +8 r 2 11 r +3 = 0. Using, say, the long division, we get 4 r 3 + 8 r 2 11 r + 3 = ( r + 3)(4 r 2 4 r + 1) = ( r + 3)(2 r 1) 2 . Thus, in addition, r = 1 / 2 is a double root of the auxiliary equation. A general solution then
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: has the Form y ( t ) = c 1 e − 3 t + c 2 e t/ 2 + c 3 te t/ 2 . 21. ±irst, we solve the corresponding homogeneous equation, y − 3 y + 7 y = 0 . Since the roots oF the auxiliary equation, r 2 − 3 r + 7 = 0, are r = 3 ± √ 9 − 28 2 = 3 ± √ 19 i 2 , a general solution to the homogeneous equation is y h ( t ) = c 1 cos √ 19 t 2 + c 2 sin √ 19 t 2 e 3 t/ 2 . 249...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online