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248_pdfsam_math 54 differential equation solutions odd

# 248_pdfsam_math 54 differential equation solutions odd - k...

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Chapter 4 7. The auxiliary equation to equation (1) on page 220 in the text, mr 2 + br + k = 0, has roots r = b ± b 2 4 mk 2 m , which are both real ( b 2 > 4 mk ) and negative because b 2 4 mk < b .L e t r 1 := b b 2 4 mk 2 m , r 2 := b + b 2 4 mk 2 m . Then a general solution to the homogeneous equation corresponding to (1) has the form y h ( t )= c 1 e r 1 t + c 2 e r 2 t . A particular solution to (1) is still given by (7) on page 221 in the text. Thus, y ( t )= c 1 e r 1 t + c 2 e r 2 t + F 0 p ( k 2 ) 2 + b 2 γ 2 sin( γt + θ ) , tan θ =( k 2 ) / ( ), is a general solution to the forced overdamped equation. 9. If a mass of m = 8 kg stretches the spring by ` =1 . 96 m, then the spring stiFness must be k = mg ` = 8 · 9 . 8 1 . 96 =40(N / m) . Substitution m =8
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Unformatted text preview: k = 40, and the external force F ( t ) = cos 2 t into the equation (23) on page 226 in the text yields 8 y + 3 y + 40 y = cos 2 t. The steady-state (a particular) solution to this equation is given in (6) and (7), page 221, that is, y p ( t ) = F ( k − mγ 2 ) 2 + b 2 γ 2 ±( k − mγ 2 ) cos γt + bγ sin γt ² = 1 [40 − (8)(2) 2 ] 2 + (3) 2 (2) 2 ±( 40 − 8(2) 2 ) cos 2 t + (3)(2) sin 2 t ² = 1 100 { 8 cos 2 t + 6 sin 2 t } = 1 10 sin(2 t + θ ) , where θ = arctan(8 / 6) ≈ . 927 . 244...
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