How to Solve Second Order Differential Equations Involving Sines and Cosines?

[tuananh3ap]

hello every body .I have some problem :
i don't know slove the equation
x"=k/m*x (1)
i don't understand result of (1) is x=Asinkt+Bcoskt

 

[CompuChip]

There are two independent solutions, because it is a second order equation. They are sine and cosine. The general solution is then a linear combination of both, as you can check by just plugging it into the differential equation.

 

[tuananh3ap]

i know this thing. But i don't know the way to slove this equation
who can help me. thank you very much

 

[CompuChip]

Well, one way is to recognise the form: if it involves just x'' and x, then it is usually something with sines and cosines, because those are the two functions you know which are (almost) their own derivatives.

If you know about complex numbers, there is a more elegant way (which also works for more general equations). That is to plug in a trial solution x = e^{\lambda t}. Then the differential equation will give you an equation for \lambda, whose roots provide the n independent solutions (for an n-th order differential equation, in this case, n = 2).
I can work out the example for you, but you better tell me whether you know about complex numbers and stuff like
e^{i\theta} = \cos\theta + i \sin\theta
before I do a lot of work for nothing