This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 216 (Section 50) Written Homework #5 – Solutions Fall 2007 1 ( 3.3–4 ) For the given differential equation 2 y 00 7 y + 3 y = 0, the characteristic equation is 2 r 2 7 r + 3 = 0 , which can be factored as follows: (2 r 1)( r 3) = 0 . Thus we have r = 1 / 2 , 3, and thus we have the following general solution: y ( x ) = c 1 e x/ 2 + c 2 e 3 x . 2 ( 3.3–6 ) For the given differential equation y 00 + 5 y + 5 y = 0, the characteristic equation is r 2 + 5 r + 5 = 0 , which can be rewritten as ( r + 5 / 2) 2 25 4 + 5 = ( r + 5 / 2) 2 5 4 = 0 = ⇒ ( r + 5 / 2) 2 = 5 4 . Thus we have complex roots r = ( 5 ± √ 5) / 2, and thus we have the following general solution: y ( x ) = e 5 x ( c 1 e √ 5 x/ 2 + c 2 e √ 5 x/ 2 ) . 3 ( 3.3–16 ) For the given differential equation y (4) + 18 y 00 + 81 y = 0, the characteristic equation is r 4 + 18 r 2 + 81 = 0 = ⇒ ( r 2 + 9) 2 = 0 = ⇒ ( r + 3 i ) 2 ( r 3 i ) 2 = 0 . Thus we have a pair of double roots r = ± 3 i , which gives the following general solution: y ( x ) = ( c 1 + c 2 x ) cos 3 x + ( c 3 + c 4 x ) sin 3 x....
View
Full
Document
This homework help was uploaded on 04/02/2008 for the course ENGR 101 taught by Professor Ringenberg during the Fall '07 term at University of Michigan.
 Fall '07
 Ringenberg

Click to edit the document details