final_solns

final_solns - APPM 2360 Final Exam: May 2, 2005 1. (25...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: APPM 2360 Final Exam: May 2, 2005 1. (25 points) (a) Find the general solution of y = 2 y + 3 y (b) Convert the equation in (a) to a first-order system of equations and write the system in matrix-vector form (c) Find the eigenvalues of the matrix in (b). How do they relate to your answer to (a)? (d) Find the general solution to the system in (b). How does this solution in particular, the individual components of the vector solution relate to your answer to (a)? Solutions: (a) y - 2 y - 3 y = 0 so try y = e rt to get char. polynomial: r 2- 2 r- 3 = ( r- 3)( r + 1) = r = 3 ,- 1. Hence general solution is y ( t ) = c 1 e 3 t + c 2 e t . (b) Define v = y , so y = v and v = y = 2 y + 3 y = 3 y + 2 v . The system is y = v v = 3 y + 2 v parenleftBigg y v parenrightBigg = parenleftBigg 0 1 3 2 parenrightBiggparenleftBigg y v parenrightBigg (c) det( A- I ) = 2- (0 + 2) + (0- 3) = 2- 2 - 3 = ( - 3)( + 1) = 0 = 3 ,- 1, which are the same are the roots of the char. poly. in (a). (d) Finding the eigenvectors: = 3 ( A- I ) = parenleftBigg- 3 1 3- 1 parenrightBigg v = parenleftBigg 1 3 parenrightBigg ; =- 1 ( A- I ) = parenleftBigg 1 1 3 3 parenrightBigg v = parenleftBigg 1- 1 parenrightBigg . Hence the general solution is parenleftBigg y v parenrightBigg = c 1 e 3 t parenleftBigg 1 3 parenrightBigg + c 1 e t parenleftBigg 1- 1 parenrightBigg The first component of the vector solution is y ( t ) = c 1 e 3 t + c 2 e t , which is the same as found in (a). The second component is v ( t ) = 3 c 1 e 3 t- c 2 e t , which is the derivative of y ( t ) (as it should be, since v was defined as y ). 1 2. (20 points) Find the general solution of y - 4 y + 4 y = f ( x ), where (a) f ( x ) = 3 e x (b) f ( x ) = 2sin(2 x ) Solutions: (a) Taking the homogeneous part: y - 4 y + 4 y =, try y = e rx and get char. polynomial: r 2- 4 r + 4 = ( r- 2) 2 = 0 r = 2 (repeated). Hence y h ( t ) = c 1 e 2 x + c 2 te 2 x . Use Undetermined Coefficients to find a particular solution: y p = Ae x y p =- Ae x and y p = Ae x . Hence y p- 4 y p + 4 y p = e x ( A + 4 A + 4 A ) = 9 Ae...
View Full Document

Page1 / 8

final_solns - APPM 2360 Final Exam: May 2, 2005 1. (25...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online