nagle_differential_equations_ISM_Part30

nagle_differential_equations_ISM_Part30 - Exercises 4.6 So...

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Unformatted text preview: Exercises 4.6 So, y p ( t ) =- 1 2 t 2 ln t- 1 4 t 2 e- 2 t + t (ln t- 1) · te- 2 t = 2 ln t- 3 4 t 2 e- 2 t , and a general solution is given by y ( t ) = 2 ln t- 3 4 t 2 e- 2 t + c 1 e- 2 t + c 2 te- 2 t . 12. The corresponding homogeneous equation is y 00 + y = 0. Its auxiliary equation has the roots r = ± i . Hence, a general solution to the homogeneous corresponding problem is given by y h = c 1 cos t + c 2 sin t. We will find a particular solution to the original equation by representing the right-hand side as a sum tan t + e 3 t- 1 = g 1 ( t ) + g 2 ( t ) , where g 1 ( t ) = tan t and g 2 ( t ) = e 3 t- 1. A particular solution to y 00 + y = g 1 ( t ) was found in Example 1, namely, y p, 1 =- (cos t ) ln | sec t + tan t | . A particular solution to y 00 + y = g 2 ( t ) can be found using the method of undetermined coefficients. We let y p, 2 = A e 3 t + B ⇒ y 00 p, 2 = 9 A e 3 t . Substituting these functions yields y 00 p, 2 + y p, 2 = ( 9 A e 3 t ) + ( A e 3 t + B ) = 10 A e 3 t + B = e 3 t- 1 . Hence, A = 1 / 10, B =- 1, and y p, 2 = (1 / 10) e 3 t- 1. By the superposition principle, y = y p, 1 + y p, 2 + y h =- (cos t ) ln | sec t + tan t | + (1 / 10) e 3 t- 1 + c 1 cos t + c 2 sin t gives a general solution to the original equation. 141 Chapter 4 14. A fundamental solution set for the corresponding homogeneous equation is y 1 ( θ ) = cos θ and y 2 ( θ ) = sin θ (see Example 1 in the text or Problem 12). Applying the method of variation of parameters, we seek a particular solution to the given equation in the form y p = v 1 y 1 + v 2 y 2 , where...
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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_Part30 - Exercises 4.6 So...

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