
Differential Equations Ordinary and Partial Differential Equations Math Forum 
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March 9th, 2016, 12:11 PM  #1 
Member Joined: Jan 2016 From: United Kingdom Posts: 35 Thanks: 0  Rearranging ODE's like algebra  please check my understanding
Hello all, I am trying to understand why we can "separate the variables and derivative parts" in ODEs. I have attached a picture of my working out, and on the page I have added numbers at the side, which I will reference now in the typed text to assist what I have done: 1. dy/dx is just the gradient of some curve. I am choosing two coordinates that are infinitely close together. Trivial, but that is the essence of differentiation so I see no issue in doing that. 2. I put x(i) and y(i) into the functions f and g because the differential equations says that for some pair of x and y, the gradient of y at that x is the same as f(x)g(y). So I am using x(i), y(i) as my pair and the (i+1) parts are merely there to measure the gradient. 3. Replacing the reciprocal of g(y) with h to make my next point clear. 4. If I graph h(y) against y and f(x) against x, what the equation says is that the area of a very thin rectangle under one graph is equal to any other thin rectangle under the other. I can pair them up so that the entire area under one graph is equal to the other. Hey presto, that's integration. Does this explain why we can get from the equation at the very top of the page to the equation at the very bottom? Last edited by skipjack; March 10th, 2016 at 04:33 AM. 
March 9th, 2016, 02:12 PM  #2 
Senior Member Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230  You may post a better−in−quality image. 
March 9th, 2016, 03:26 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra 
A better way is as follows: $$\begin{aligned} {\mathrm d y \over \mathrm d x} &= f(x)g(y) = {f(x) \over h(y)} \\ h(y){\mathrm d y \over \mathrm d x} &= f(x)) \\ \int h(y){\mathrm d y \over \mathrm d x} \, \mathrm d x &= \int f(x)\,\mathrm d x \\ \int h(y)\, \mathrm d y &= \int f(x)\,\mathrm d x \end{aligned}$$ Where the last line is an application of a $u$ substitution (or its reverse), and can be shown using the Fundamental Theorem of Calculus and the chain rule. 
March 10th, 2016, 04:23 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Another way of looking at it the derivative, dy/dx, is NOT a fraction, but it is defined as the limit of a fraction. Because of that you can show that it can be treated like a fraction (as in "the chain rule", (dy/dx)(dx/dt)= dy/dt) by going back before the limit, using the fraction properties, then taking the limit.

March 10th, 2016, 08:14 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra 
That's not a valid detivation of the chain rule for technical reasons.

March 10th, 2016, 09:31 AM  #6 
Member Joined: Jan 2016 From: United Kingdom Posts: 35 Thanks: 0 
v8archie  can you please explain where my ideas run into technical issues?

March 10th, 2016, 10:03 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra 
I haven't read your work in detail, but I'll take a look tonight.

March 13th, 2016, 06:39 AM  #8 
Member Joined: Jan 2016 From: United Kingdom Posts: 35 Thanks: 0 
v8archie  did you get a chance to look at it?


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algebra, check, ode, rearranging, understanding 
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