What are differential equations good for?

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SUMMARY

Differential equations are essential for modeling real-world phenomena, such as the motion of objects connected to springs, as illustrated by Hooke's Law and Newton's Laws. The equation $$\ddot x(t) + \frac{k}{m}x(t) = 0$$ represents the relationship between acceleration and position for a mass-spring system. Understanding these equations requires knowledge of derivatives and their applications in physics. The discussion emphasizes the importance of connecting mathematical concepts to physical scenarios for better comprehension.

PREREQUISITES
  • Understanding of basic calculus, including differentiation and integration.
  • Familiarity with Newton's Laws of motion.
  • Knowledge of Hooke's Law and its implications in physics.
  • Basic understanding of functions and their derivatives.
NEXT STEPS
  • Explore the applications of differential equations in physics, particularly in mechanical systems.
  • Study the derivation and solutions of second-order differential equations.
  • Learn about partial differential equations and their applications in modeling complex systems.
  • Investigate numerical methods for solving differential equations, such as the Runge-Kutta method.
USEFUL FOR

Students of physics and mathematics, engineers involved in mechanical systems, and anyone interested in understanding the practical applications of differential equations in modeling dynamic systems.

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When I learned differential calculus, my instructor spoke only in terms of x's & y's. I could differentiate, but had absolutely no idea what I was doing or how it applied.

My mentor pulled me aside, gave me a function, f(x)=x^2, & made note that the function was modeling a vehicle's position in space over time. He then asked me what the vehicle's velocity was at some time, t. f'(x)=2x. In other words, the vehicle was accelerating at a rate of 2mph per hour. Okay, now I get it. I see how it applies. I get why we use it. He did the same thing for integrals.My mentor is no longer around & now I am back in the same position - I am solving differential equations but have absolutely no idea what I'm doing. Can we use the vehicle's position over time scenario?

A diff eq involves a function & it's derivative, right?

So, x^2 + 2x = ?

I'm not even sure if I understand that operation -why in the world would you add a function & it's derivative in the first place? What should we include on the other side of the equation & what would that mean?
 
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Differential equations arise when you begin to model things happening in the real world. To model something means to use concepts from physics together with mathematics, and then create a number of equations that describe the behavior of the system you are interested in studying.

A very simple example is the problem of an object with a given mass ##m## connected to a spring. The idea here, is that we are interested in figuring out, how this object moves, or how we can describe it's motion in terms of some equations.

Snapshot.jpg


From physics, you would then know something about this system:
1) You would know that a if the spring is stretched or compressed, it will produce a force in the opposite direction. That is - if you push it pushes back, if you stretch it pulls back. This principle can be written as an equation and is known as Hooke's Law. Namely -
$$ F = -k x $$
Here the constant ##k## is a number that relates the displacement from its normal length x to the force it produces.
Well that's all nice and well, but we also got to look at the mass. From Newtons Laws we know that the acceleration of an object is proportional to the total force applied on it.
$$ma = F_{net}$$
In our case, we are so lucky that the total force that pushes on the object is only the force coming from the spring. And we know what the spring force is! So all we got to do now is insert the Hooke's Law equation into the Newton's Law equation, and we get
$$ma = -kx$$
So what we have here is an equation that relates the acceleration of the object, to its position. It's important to understand at this point, that x, is some function that at any given time, tells us where the mass is. If x = 0, then it means that the mass is at the position, where the spring is neither stretched nor compressed.

The last idea is then to use the knowledge about the car moving and stuff like that, which allows you to say that the acceleration, a, is equal to the second derivative of it's position - ##\ddot x(t)##. When this final piece is put into the equation, you can shuffle it around and you get
$$ \ddot x(t) + \frac{k}{m}x(t) = 0 $$
And that is a differential equation. In order to figure out what the actual function x(t) looks like, we would then use all sorts of math belonging to the world of differential equations. If the values k and m are just right, you end up with the equation you described :)

The above example is a very simple one, and you quickly get fantastically complex equations like
$$ B\left( \frac{\partial^4f}{\partial x^4} + 2\frac{\partial^4f}{\partial x^2\partial y^2} + \frac{\partial^4f}{\partial y^4} \right) + \bar m \frac{\partial^2f}{\partial t^2} = p(x,y,t) $$
The above equation is the differential equation (partial differential equation actually), that tells you what happens when you have a plate and you vibrate it. Figuring out what happens means figuring out the types of functions, ##f(x,y,t)##, that can be plugged into that differential equation and satisfy it.

Hope this shed some light :)
 
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