What exactly is a 2nd order differential equation?

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SUMMARY

A second order differential equation (DE) describes the rate of change of a rate of change, exemplified by the equation m(d²x/dt²) = -mg, which models the motion of an object under the influence of gravity. This mathematical representation is foundational in classical physics, particularly in mechanics and electrical circuits involving inductors, resistors, and capacitors. The discussion highlights practical applications of second order DEs, including the motion of a ball thrown upward and the behavior of RLC circuits. Understanding these concepts is essential for applying differential equations in real-world scenarios.

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Venomily
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A first order DE models the rate of change, e.g. when decay is proportional to time we have the DE: dM/dt = -K.M; this is describing that rate of change mathematically. Am I correct in saying that a 2nd order DE describes the rate of rate of change?

Also, can anyone explain any application of 2nd order DEs to me? I understand it mathematically, but I am interested in how it works in practice like that decay example I pointed out above, hence this post.
 
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Venomily said:
Also, can anyone explain any application of 2nd order DEs to me?
Classical physics is chock full of second order ODEs. F=ma, for example.
 
D H said:
Classical physics is chock full of second order ODEs. F=ma, for example.

Not to mention electrical circuits with an inductor, resistor, and capacitor.
 
D H said:
Classical physics is chock full of second order ODEs. F=ma, for example.

Mark44 said:
Not to mention electrical circuits with an inductor, resistor, and capacitor.

Thanks, but can you go through an example with me? actually point out a real life application (which you guys did) but also deriving a 2nd order DE to model it step by step.
 
If you throw a ball directly upward with initial speed 10 m/s, the basic physics law is "force= mass times acceleration". In this case, the only force (neglecting air resistance) is gravity: -mg. Since acceleration is the second derivative of the position function, taking x to be the height above the ground at time t, we have the differential equation
m\frac{d^2x}{dt^2}= -mg
with initial conditions x(0)= 0, x'(0)= 10.
 

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