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final_solns

# final_solns - APPM 2360 Final Exam May 2 2005 1(25 points(a...

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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) = 0 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 - λI ) = λ 2 - (0 + 2) λ + (0 - 3) = λ 2 - 2 λ - 3 = ( λ - 3)( λ + 1) = 0 λ = 3 , - 1, which are the same are the roots of the char. poly. in (a). (d) Finding the eigenvectors: λ = 3 ( A - λI ) = parenleftBigg - 3 1 3 - 1 parenrightBigg v = parenleftBigg 1 3 parenrightBigg ; λ = - 1 ( A - λI ) = 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

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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 ) = 2 sin(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 x = 3 e x A = 1 3 .
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final_solns - APPM 2360 Final Exam May 2 2005 1(25 points(a...

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