183_pdfsam_math 54 differential equation solutions odd

# 183_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.3 So, α = 1, β = 2, and a general solution is given by y ( t )= c 1 e t cos 2 t + c 2 e t sin 2 t. 15. First, we fnd the roots o± the auxiliary equation. r 2 +10 r +41=0 r = 10 ± p 10 2 4(1)(41) 2 = 5 ± 4 i. These are complex numbers with α = 5and β = 4. Hence, a general solution to the given di²erential equation is y ( t )= c 1 e 5 t cos 4 t + c 2 e 5 t sin 4 t. 17. The auxiliary equation in this problem, r 2 r + 7 = 0, has the roots r = 1 ± p 1 2 4(1)(7) 2 = 1 ± 27 2 = 1 2 ± 3 3 2 i. There±ore, a general solution is y ( t )= c 1 e t/ 2 cos 3 3 2 t + c 2 e t/ 2 sin 3 3 2 t . 19. The auxiliary equation, r 3 + r 2 +3 r 5 = 0, is a cubic equation. Since any cubic equation has a real root, frst we examine the divisors o± the ±ree coeﬃcient, 5, to fnd integer real roots (i±
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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