This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full 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 firstorder system of equations and write the system in matrixvector 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
 Summer '07
 WILLIAMHEUETT

Click to edit the document details