Move 'dx' from 'dy/dx' to the other side, then integrate both sides

  • Thread starter Thread starter knghcm
  • Start date Start date
  • Tags Tags
    Dx Integrate
knghcm
Messages
2
Reaction score
0
I've been reading in my engineering textbooks and came across a frequent equation manipulation that involves multiplying/bringing the dx term of dy/dx to the other side of the equation, and then integrate both sides. I don't know what technique this is and I can't find it in my Stewart's Calculus either. Please help me find out what this is and how I can read about it more. I've heard that the individual dy and dx terms are called infinitesimals but when I googled infinitesimal calculus, the results all went way over my head (wtf is hyperreal?)

I really need your help because I'm starting to get lost on the math on my engineering classes. Thank you.

Edit: okay I've just done some googling and it turned out to be a simple differential equation problem. Since I have not taken any differential equation yet, what is a good way to quickly gain ground on this topic so that it could be useful for other classes while not getting bogged down so much that it's like another class of its own.
 
Last edited:
Physics news on Phys.org


It is an example of bad mathematics: It works for engineering/physics equations, but it is not a proper way to handle derivatives.
 


It's not an example of bad mathematics... There is a formal theory of differential forms, and bringing the dx to the other side is just simply used to express a relation between the differentials... Of course, without resorting to differentials, you can just interpret it as a change of variable, but that's not necessarily any simpler...

To the OP: You don't need to know anything about hyperreals :) That's just a separable differential equation. Every class of differential equations has its tricks ;)

If your equation is of the form: f(y)\frac{dy}{dx}=g(x), just integrate both sides and use a change of variables to obtain an implicit expression for y:
f(y)\frac{dy}{dx}=g(x) \Rightarrow \int_{x_0}^xf(y)\frac{dy}{dx}dx=\int_{x_0}^xg(x)dx\Rightarrow \int_{y_0}^yf(y)dy=\int_{x_0}^xg(x)dx
(Technically, the variable of integration can not simultaneously be its upper limit, but just switch the y's and x's inside the integrals to y' and x' or something, to distinguish them from the limits)
 


Thank you so much for your answers. Bad mathematics or not, where I can go to to learn more about this technique? It's coming up a lot in many engineering "proofs" I've come across so I need to know it pretty well.

Edit: I just picked up the Boyce and DiPrima ODE book from my college library, and they are called separable equations (section 2.2 in the 9th edition). I will definitely spend much time on this. If there's any other resources that I need to be aware of please let me know.
 
Last edited:


knghcm said:
Thank you so much for your answers. Bad mathematics or not, where I can go to to learn more about this technique? It's coming up a lot in many engineering "proofs" I've come across so I need to know it pretty well.

Edit: I just picked up the Boyce and DiPrima ODE book from my college library, and they are called separable equations (section 2.2 in the 9th edition). I will definitely spend much time on this. If there's any other resources that I need to be aware of please let me know.

I don't know anything about that particular book, but any introductory ODE textbook should suit you fine. There's not really that much to know about separable equations, just a couple of tricks to recognize if a certain equation can be made separable under a certain substitution, or rearranging the terms, etc. Other than that, it's just integrating both sides ;)

Also very common and of utmost importance are linear differential equations, and there's also a simple trick you can use to solve them. Another large class of ODEs are exact equations, or ones that can be reduced to exact. Depending on your needs, you may need to solve systems of linear differential equations, or differential equations with higher order derivatives. It's an interesting subject, just pick up a book and read, or maybe watch MIT's lecture series, although he starts off by mentioning that he expects students to already know how to solve separable equations ;)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top