soln_hw5 - Solutions to Homework 5 Section 3.3 1 Suppose...

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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 / 2 sin(2 t ). As t → ∞ , y continues to oscillate between - 1 and 1. 0 1 2 3 4 5 6 7 8 9 10 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 t y y = 1/2 sin(t) 24. a. The characteristic equation is 5 r 2 + 2 r + 7 = 0, with roots r = - 1 / 5 ± i 34 / 5. The solution
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