nagle_differential_equations_ISM_Part28

nagle_differential_equations_ISM_Part28 - Exercises 4.5 and...

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Unformatted text preview: Exercises 4.5 and a general solution is y ( t ) = y p ( t ) + y h ( t ) = t 2- 4 t + 7- e t 4 + c 1 e- t + c 2 te- t . Next, we satisfy the initial conditions. 0 = y (0) = 7- 1 / 4 + c 1 2 = y (0) =- 4- 1 / 4- c 1 + c 2 c 1 =- 27 / 4 c 2 = 25 / 4 + c 1 =- 1 / 2 . Therefore, the solution to the given initial value problem is y ( t ) = t 2- 4 t + 7- e t 4- 27 e- t 4- te- t 2 . 32. For the nonhomogeneous term, e 2 t + te 2 t + t 2 e 2 t = ( 1 + t + t 2 ) e 2 t , a particular solution has the form y p ( t ) = t s ( A + A 1 t + A 2 t 2 ) e 2 t . Since r = 2 is not a root of the auxiliary equation, r 2- 1 = 0, we choose s = 0. 34. Neither r = i nor r = 2 i is a root of the auxiliary equation, which is r 2 + 5 r + 6 = 0. Thus, by the superposition principle, y p ( t ) = A cos t + B sin t + C cos 2 t + D sin 2 t. 36. Since the auxiliary equation, r 2- 4 r + 4 = ( r- 2) 2 = 0 has a double root r = 2 and the nonhomogeneous term can be written as t 2 e 2 t- e 2 t = ( t 2- 1 ) e 2 t , a particular solution to the given equation has the form y p ( t ) = t 2 ( A 2 t 2 + A 1 t + A ) e 2 t . 38. Since, by inspection, r = i is not a root of the auxiliary equation, which is r 4- 5 r 2 +4 = 0, we look for a particular solution of the form y p ( t ) = A cos t + B sin t. 131 Chapter 4 Differentiating y p ( t ) four times, we get y p ( t ) =- A sin t + B cos t, y 00 p ( t ) =- A cos t- B sin t, y 000 p ( t ) = A sin t- B cos t, y (4) p ( t ) = A cos t + B sin t....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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nagle_differential_equations_ISM_Part28 - Exercises 4.5 and...

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