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Homework 2, due September 11, 2013.

Posted: Fri Sep 06, 2013 9:02 am
by goodwine
Reading: Chapter 2 of the course text.

Exercises: 1.8 (odd ones only), 1.12 (even ones only), 2.2-2.6, 2.8-2.9.

There will be no extensions granted because there is an exam on Friday.

Re: Homework 2, due September 11, 2013.

Posted: Mon Sep 09, 2013 11:05 pm
by sbrill
For question 1.12 #10. Is x(t)=t+1/t+5t/(2ln(t)) or is x(t)=t+1/t+(5t/2)(ln(t))? Basically, in the last term, is ln(t) in the numerator or the denominator?

Re: Homework 2, due September 11, 2013.

Posted: Tue Sep 10, 2013 5:45 am
by goodwine
sbrill wrote:For question 1.12 #10. Is x(t)=t+1/t+5t/(2ln(t)) or is x(t)=t+1/t+(5t/2)(ln(t))? Basically, in the last term, is ln(t) in the numerator or the denominator?
It's in the denominator.

Whenever there is an ambiguity like this it's always permissible to put on your homework an indication which way you assumed it was. (You can always ask too).

Re: Homework 2, due September 11, 2013.

Posted: Tue Sep 10, 2013 9:33 am
by bgerig1
For a function to be considered a solution to a differential equation, does the function have to be a solution for all values of the independent variable? Or can you consider a function a solution if you specify what values of the independent variable must be plugged in to the function for the function to solve the differential equation? This relates to problem 1.12.

Re: Homework 2, due September 11, 2013.

Posted: Tue Sep 10, 2013 10:07 am
by goodwine
bgerig1 wrote:For a function to be considered a solution to a differential equation, does the function have to be a solution for all values of the independent variable? Or can you consider a function a solution if you specify what values of the independent variable must be plugged in to the function for the function to solve the differential equation? This relates to problem 1.12.
That's a good question. If a problem doesn't limit the domain of the independent variable, then implicitly the problem is saying it must hold for all values. If the problem limits the domain for the independent variable, then it only has to hold for that limited set of values.

This is usually not stated because it is sort of obviously from the context of the problem. However, it can also be one of those things that becomes very confusing because it is not explicitly stated.