Re: Exp-Trig Manipulation
- To: mathgroup at smc.vnet.net
- Subject: [mg59388] Re: Exp-Trig Manipulation
- From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
- Date: Mon, 8 Aug 2005 03:34:35 -0400 (EDT)
- Organization: Uni Leipzig
- References: <dd4ejd$hls$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi, e1 = E^((2 - I)*x)*C[1] + E^((2 + I)*x)*C[2] + E^((3 - 4*I)*x)*C[3] + E^((3 + 4*I)*x)*C[4] // ComplexExpand ?? Regards Jens "Daniele Lupo" <danwolf80_no_spam_please_ at libero.it> schrieb im Newsbeitrag news:dd4ejd$hls$1 at smc.vnet.net... | Hi to everyone. | | I'd like to know how I can convert this expression | | E^((2 - I)*x)*C[1] + E^((2 + I)*x)*C[2] + E^((3 - 4*I)*x)*C[3] + E^((3 + | 4*I)*x)*C[4] | | in its equivalent form | | E^(2*x)*C[2]*Cos[x] + E^(3*x)*C[4]*Cos[4*x] + E^(2*x)*C[1]*Sin[x] + | E^(3*x)*C[3]*Sin[4*x] | | | I've obtained them while resolving a differential equation. I've tried to | solve this: | | car = y''''[x] - 10*y'''[x] + 54*y''[x] - 130*y'[x] + 125*y[x] == 0; | | In two different ways: first, working with characteristic polynomial: | | ------------------- | | (* Conversion from differential equation to characteristic polynomial *) | | pol = car /. {Derivative[n_][y][x] -> ë^n, y[x] -> 1}; | | (* Solutiof of c.p. *) | | sol = Solve[pol, ë]; | | (* Mapping solutions in a linear combination of exponentials *) | | solution1 = Plus @@ MapIndexed[C[#2[[1]]]*Exp[x*#1] & , ë /. %] | | ------------------- | | | While I've obtained second solution using DSolve: | | | ------------------- | | solution2 = y[x] /. DSolve[car, y[x], x][[1]] | | ------------------- | | So, if I did not wrong something, these two solutions must be equivalent, | but I can't find a way to trasform solution1 to solution2: I know that | there can be a problem in conversion of C[n] coefficients during | transformation, but I don't care it. I'd like instead to convert in the | right way exponentials of first method in right product of Cos, Sin, Exp of | the solution obtained with DSolve. | | Thanks for answers | | Daniele |