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ODEs and Integration with the Chain Rule

Date: 07/04/98 at 08:36:07
From: Toby Smith
Subject: Integration

You guys are being extremely helpful this year. I don't know how I
would be doing my distance ed. course without you. Thanks!

Anyway, I have more integration problems for you.

Q1) What is an ordinary differential equation? Can you give me an 
example of one, and an example of a DE that isn't ordinary?

Q2) How do I integrate:

   f'(x) = (x-1)^4  

I tried applying the chain rule as in differentiation, and I get:

   f'(x) = (x-1)^4
   f(x) = (1/5)*(x-1)^5  * ((1/2)*(x^2) - x) + C

But my text gives just:

   f(x) = (1/5)*(x-1)^5 + C

Is the text correct? Where did I go wrong?
What is the correct method for integrating DE's of the form 
dy/dx = (f(x))^n?

Again, thanks very much.

Date: 07/04/98 at 08:59:46
From: Doctor Jerry
Subject: Re: Integration

Hi Toby,

Question 1:

There are ordinary differential equations and partial differential 
equations. ODEs are those involving one independent variable (for 
example x) and one dependent variable (for example y). An ODE is an 
equation involving x, y, y', y'', or other higher-order derivatives.  
y'' - 6y' + y = sin(x) is a second order ordinary differential 

Partial differential equations are those involving one dependent 
variable and several independent variables. For example, if z is a 
dependent variable of the independent variables x and y, then a PDE 
is an equation involving z, x, y, and various orders of partial 
derivatives of z with respect to x and y. z_{xx} + z_{yy} = x^2 + y^2  
is an example of a PDE. The expression z_{xx} means the second partial 
of z with respect to x.

Question 2:

You want to find a function f whose derivative is (x-1)^4. You are 
correct in thinking that the chain rule is involved. You want to 
figure out an expression whose derivative is (x-1)^4. From  your 
general knowledge of the chain rule, you would guess that it must have 
been something like (x-1)^5. The derivative of this is 5(x-1)^4. So, 
you say to yourself, it must be (1/5)(x-1)^5. The factor of (1/5) is 
there to "catch" the 5 as it comes down. Also, of course, it could 
have been (1/5)(x-1)^5+c, where c is any constant. Because, when c is 
differentiated, it contributes only 0.

For your last question, there is no universally applicable formula or 
procedure for integrating expressions of the form (f(x))^n. This is 
why integration is not so easy as differentiation. There are quite a 
few different methods. Moreover, some expressions of this form cannot 
be integrated at all, not in "closed form."

- Doctor Jerry, The Math Forum
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Associated Topics:
High School Calculus

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