Homework 2, due September 11, 2013.

Due at the beginning of class on Wednesday, September 11, 2013.
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goodwine
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Homework 2, due September 11, 2013.

Post 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.
Bill Goodwine, 376 Fitzpatrick
sbrill

Re: Homework 2, due September 11, 2013.

Post 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?
goodwine
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Re: Homework 2, due September 11, 2013.

Post 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).
Bill Goodwine, 376 Fitzpatrick
bgerig1

Re: Homework 2, due September 11, 2013.

Post 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.
goodwine
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Re: Homework 2, due September 11, 2013.

Post 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.
Bill Goodwine, 376 Fitzpatrick
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