ODEs and Integration with the Chain RuleDate: 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 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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