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### ODEs and Integration with the Chain Rule

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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.
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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
equation.

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
Check out our web site! http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus

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