Mar 1, 2006 #1 S sky_tm Junior Member level 1 Joined Feb 15, 2006 Messages 15 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,262 Higher Derivatives \[\frac{{d^3 y}}{{dx^3 }} - 4\frac{{d^2 y}}{{dx^2 }} + 16\frac{{dy}}{{dx}} = 0\] Solve.
Higher Derivatives \[\frac{{d^3 y}}{{dx^3 }} - 4\frac{{d^2 y}}{{dx^2 }} + 16\frac{{dy}}{{dx}} = 0\] Solve.
Mar 1, 2006 #2 H Hughes Advanced Member level 3 Joined Jun 10, 2003 Messages 715 Helped 113 Reputation 226 Reaction score 26 Trophy points 1,298 Activity points 5,984 Re: Higher Derivatives λ³-4λ²+16λ=0 λ1 = 0, λ2 = 2+4\[\sqrt 3\]i, λ3 = 2-4\[\sqrt 3\]i y = C1 + C2 exp(2x) cos(4\[\sqrt 3\]x) + C3 exp(2x) sin(4\[\sqrt 3\]x)
Re: Higher Derivatives λ³-4λ²+16λ=0 λ1 = 0, λ2 = 2+4\[\sqrt 3\]i, λ3 = 2-4\[\sqrt 3\]i y = C1 + C2 exp(2x) cos(4\[\sqrt 3\]x) + C3 exp(2x) sin(4\[\sqrt 3\]x)
Mar 3, 2006 #3 S sky_tm Junior Member level 1 Joined Feb 15, 2006 Messages 15 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,262 Re: Higher Derivatives thanks for the ans. but its different from mine... so can you help me find out what went wrong? \[\lambda = 0\] \[\lambda = 2 + 2j\sqrt 3\] \[\lambda = 2 - 2j\sqrt 3\] and btw how u got the exp, C1 C2 C3 ?
Re: Higher Derivatives thanks for the ans. but its different from mine... so can you help me find out what went wrong? \[\lambda = 0\] \[\lambda = 2 + 2j\sqrt 3\] \[\lambda = 2 - 2j\sqrt 3\] and btw how u got the exp, C1 C2 C3 ?
Mar 3, 2006 #4 H Hughes Advanced Member level 3 Joined Jun 10, 2003 Messages 715 Helped 113 Reputation 226 Reaction score 26 Trophy points 1,298 Activity points 5,984 Re: Higher Derivatives sky_tm said: thanks for the ans. but its different from mine... so can you help me find out what went wrong? \[\lambda = 0\] \[\lambda = 2 + 2j\sqrt 3\] \[\lambda = 2 - 2j\sqrt 3\] and btw how u got the exp, C1 C2 C3 ? Click to expand... Sorry, it's my mistake. It should be 2 other than 4. As for exp, you can consider it a formula. C1, C2, C3 are arbitary consts.
Re: Higher Derivatives sky_tm said: thanks for the ans. but its different from mine... so can you help me find out what went wrong? \[\lambda = 0\] \[\lambda = 2 + 2j\sqrt 3\] \[\lambda = 2 - 2j\sqrt 3\] and btw how u got the exp, C1 C2 C3 ? Click to expand... Sorry, it's my mistake. It should be 2 other than 4. As for exp, you can consider it a formula. C1, C2, C3 are arbitary consts.
Mar 9, 2006 #5 D dreamcard Member level 2 Joined Sep 11, 2004 Messages 49 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 309 Higher Derivatives this can also be solved by using Laplace transformation. the zero input response of the equation
Higher Derivatives this can also be solved by using Laplace transformation. the zero input response of the equation