hwk_05( http-::math.uakron.edu:faculty:montero:diff_eq:hwk_sols:hwk_05_sols.pdf)

Hwk_05( http-::math.uakron.edu:faculty:montero:diff_eq:hwk_sols:hwk_05_sols.pdf)

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Solutions to Homework 5, Introduction to Diferential Equations, 3450:335-003, Dr. Montero, Spring 2009 Problem 1. Find a particular solution to the diferential equation y ′′ - 4 y = 48 t 3 . Solution: First we look at the polynomial associated to the right hand side: λ 2 - 4 = ( λ - 2)( λ + 2) . The roots o± this polynomial are ± 2. Next we observe that annihilator ±or the right hand side is D 4 , that has the associated polynomial λ 4 . This polynomial has only one root, 0, with multiplicity 4, and this is, o± course, diferent ±rom ± 2. Hence we seek a particular solution o± the ±orm y ( t ) = α 1 + α 2 t + α 3 t 3 + α 4 t 3 . We ²nd next y ( t ) and y ′′ ( t ) to obtain y ( t ) = α 2 + 2 α 3 t + 3 α 4 t 2 and y ′′ ( t ) = 2 α 3 + 6 α 4 t. We plug this into the equation, and group together the terms with t 0 , t 1 , t 2 and t 3 , to obtain 2 α 3 + 6 α 4 t - 4( α 1 + α 2 t + α 3 t 2 + α 4 t 3 ) = (2 α 3 - 4 α 1 ) + (6 α 4 - 4 α 2 ) t - 4 α 3 t 2 - 4 α 4 t 3 = - 48 t 3 . We solve ±rom here to obtain - 4 α 4 = - 48 so α 4 = 12 , α 3 = 0 , 6 α 4 - 4 α 2 = 0 , so α 2 = 18 , and 2 α 3 - 4 α 1 = 0 so α 1 = 0 .
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Problem 2. Find a particular solution of y ′′ + 6 y + 8 y = 13 te 5 t . Solution: The annihilator for the right hand side is ( D - 5) 2 , that has associated polynomial ( λ - 5) 2 , of if one prefers, λ 2 - 10 λ + 25 . This also says that the function g ( t ) = 13 te 5 t satis±es the equation y ′′ - 10 y + 25 y = 0 . We will make use of this fact later to simplify our computations. We also note that the roots of λ 2 - 10 λ + 25 . are 5 and 5 (or 5 with multiplicity 2). On the other hand, the polynomial for the left hand side is λ 2 + 6 λ + 8 = ( λ + 4)( λ + 2) , that has roots - 2 and - 4. Neither of these is 5, so we seek a particular solution of the form y p ( t ) = α 1 e 5 t + α 2 te 5 t = ( α 1 + α 2 t ) e 5 t . Notice now that, due to the way this function was built, y p satis±es y ′′ p - 10 y p + 25 y p = 0 . We must plug y p into the equation y ′′ + 6 y + 8 y = 13 te 5 t . To do this we take advantage of the fact that y ′′ p - 10 y p + 25 y p = 0 . as follows: y ′′ + 6 y + 8 y = ( y ′′ - 10 y + 25 y ) + 16 y + 17 y,
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so y ′′ p + 6 y p + 8 y p = ( y ′′ p - 10 y p + 25 y p ) + 16 y p + 17 y p = 16 y p + 17 y p . Because of this, when we plug y p into y ′′ + 6 y + 8 y = 13 te 5 t the equation reduces to 16 y p + 17 y p = 13 te 5 t .
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Hwk_05( http-::math.uakron.edu:faculty:montero:diff_eq:hwk_sols:hwk_05_sols.pdf)

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