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Unformatted text preview: Solutions to Homework 5 Section 3.3 1. Suppose that αf ( t ) + βg ( t ) = 0, that is, α ( t 2 + 5 t ) + β ( t 2 5 t ) = 0 on some interval I . Then ( α + β ) t 2 + 5( α β ) t = 0, for all t ∈ I . Since a a quadratic has at most two roots, we must have α + β = 0 and α β = 0. The only solution is α = β = 0. Hence the two functions are linearly independent. 4. Obviously, f ( x ) = eg ( x ) for all real numbers x . Hence the functions are linearly dependent. 5. Here f ( x ) = 3 g ( x ) for all real numbers x . Hence the functions are linearly dependent. Section 3.4 17. The characteristic equation is r 2 + 4 = 0, with roots r = ± 2 i . Hence the general solution is y = c 1 cos(2 t ) + c 2 sin(2 t ). Its derivative is y ′ = 2 c 1 sin(2 t ) + 2 c 2 cos(2 t ). Based on the first condition, y (0) = 0, we require that c 1 = 0. In order to satisfy the condition y ′ (0) = 1, we find that 2 c 2 = 1. The constants are c 1 = 0 and c 2 = 1 / 2. Hence the specific solution is y ( t ) = 1 / 2sin(2 t ). As t → ∞ , y continues to oscillate between 1 and 1....
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This note was uploaded on 10/23/2009 for the course MATH 22B taught by Professor Hunter during the Winter '08 term at UC Davis.
 Winter '08
 Hunter
 Differential Equations, Equations

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