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Unformatted text preview: Chapter 4 16. In this homogeneous CauchyEuler equation a = 1, b = 3, and c = 6. Therefore, the corresponding auxiliary equation, ar 2 + ( b a ) r + c = r 2 4 r + 6 = ( r 2) 2 + 2 = 0 has complex roots r = 2 2 i with = 2, = 2. According to (8) in the text, the functions y 1 ( t ) = t 2 cos 2 ln t , y 2 ( t ) = t 2 sin 2 ln t are two linearly independent solutions to the given homogeneous equation. Thus, a general solution is given by y = c 1 y 1 + c 2 y 2 = t 2 h c 1 cos 2 ln t + c 2 sin 2 ln t i . 18. The substitution y = t r leads the characteristic equation (see (7)) r ( r 1) + 3 r + 5 = 0 r 2 + 2 r + 5 = ( r + 1) 2 + 4 = 0 . Solving yields r = 1 2 i. Thus, the roots are complex numbers i with = 1, = 2. According to (8) in the text, the functions y 1 ( t ) = t 1 cos(2 ln t ) , y 2 ( t ) = t 1 sin(2 ln t ) are two linearly independent solutions to the given homogeneous equation. Thus, a general solution is given by y = c 1 y 1 + c 2 y 2 = t 1 [ c 1 cos(2 ln t ) + c 2 sin(2 ln t )] . 20. First, we find a general solution to the given CauchyEuler equation. Substitution y = t r leads to the characteristic equation r ( r 1) + 7 r + 5 = r 2 + 6 r + 5 = 0 r = 1 , 5 . Thus, y = c 1 t 1 + c 2 t 5 is a general solution. We now find constants c 1 and c 2 such that the initial conditions are satisfied. 1 = y (1) = c 1 + c 2 , 13 = y (1) = c 1 5 c 2 c 1 = 2 , c 2 = 3 and, therefore, y = 2 t 1 3 t 5 is the solution to the given initial value problem....
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 Spring '08
 MAZMANI
 Equations

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