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Hi, I want to ask how to solve this equation in this way?
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The difference between the previous differential equation and this one is that the first one had a second derivative in it and this one a fourth derivative. So the characteristic equation changes from [itex]r^2=\lambda_i^2[/itex] to [itex]r^4=\mu_i^4[/itex]. This has the solutions [itex]r=\{\mu_i,\mu_i,i \mu_i, i\mu_i \}[/itex] as a result the general solution will become [itex]\psi(x)=A e^{\mu x}+B e^{\mu x}+C e^{ i \mu x}+D e^{i \mu x}[/itex]. The exponents with i in their powers can be written as cosine and sine again and the real exponents can be written as cosh and sinh.For this differential equation, the characteristic equation is [itex]r^2= \lambda_i[/itex] and has characteristic roots [itex]r= \pm\lambda_i i[/itex] (that second "i" is the imaginary unit). That, in turn means that the general solution to the differential equation is [itex]\psi_i(x)= Ae^{i\lambda_i x}+ Ce^{i\lambda_i x}= Ccos(\lambda_i x)+ D sin(\lambda_i x)[/itex] where A, B, C, and D are constants. Do you understand how to convert from the complex exponential to sine and cosine? It is based on [itex]e^{ix}= cos(x)+ i sin(x)[/itex].
That is all correct so far. You can now solve equation (5) for C3 and then plug it in to solve for C4. After that you can solve C1 and C2. It doesn't look like it's going to be pretty though, I'll have a look at it later to see if you can simplify it.
We don't. If [itex]cosh(\mu)cos(\mu)= 1[/itex], so there are nontrivial solutions, there will be an infinite number of such nontrivial functions.thanks alot,
if [itex]C_1\ne 0[/itex]
then how we find [itex]C_1[/itex] ?
[/URL]thank you very much... I would like to give you this gift for helping me
http://blog.doctissimo.fr/php/blog/un_avenir_heureux/images/bouquet%20de%20fleur.gif [Broken]