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

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Homework Help Overview

The discussion revolves around solving a second-order differential equation of the form x'' = k/m * x, which is related to the behavior of oscillatory systems. The subject area includes differential equations, particularly those involving trigonometric functions like sine and cosine.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the solutions, noting that sine and cosine functions are independent solutions to the equation. There is an inquiry about the method to derive the general solution, with some participants suggesting the recognition of the equation's form as indicative of oscillatory behavior.

Discussion Status

The conversation is ongoing, with some participants providing insights into the general solution and the use of complex numbers as a potential method for solving the equation. There is no explicit consensus, but a productive direction is emerging regarding the exploration of different solution methods.

Contextual Notes

Participants express uncertainty about their understanding of complex numbers and their relevance to solving the equation, indicating a potential gap in foundational knowledge that may affect the discussion.

tuananh3ap
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hello every body .I have some problem :
i don't know slove the equation
[tex]x"=k/m*x[/tex] (1)
i don't understand result of (1) is [tex]x=Asinkt+Bcoskt[/tex]
 
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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.
 
i know this thing. But i don't know the way to slove this equation
who can help me. thank you very much
 
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 [itex]x = e^{\lambda t}[/itex]. Then the differential equation will give you an equation for [itex]\lambda[/itex], 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
[tex]e^{i\theta} = \cos\theta + i \sin\theta[/tex]
before I do a lot of work for nothing :smile:
 

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