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Ask and answer questions about the homework here. If asking a new question please start a new section at the bottom of the page. | Ask and answer questions about the homework here. If asking a new question please start a new section at the bottom of the page. | ||
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| + | I was doing supplemental problem A and found: | ||
| + | <math>e^{-2t}(2A-2At^2+4At-3)=0.</math> | ||
| + | So if t=0 then A=3/2 | ||
| + | And: | ||
| + | <math>e^{-2t}(-4Bt+4B-3)=0.</math> | ||
| + | So if t=0 then B=4/3 | ||
| + | But with the variable t I wasn't sure if these answers were enough for A and B or if I needed to find more possibilities, which I had no clue how to do. | ||
| + | |||
| + | :ANSWER: You need to find values of A and B that work for all t (if there are any). Also, I think you made a mistake when taking the derivatives. Note that the derivative of <math>e^{-2t}</math> is <math>-2 e^{-2t}</math>. | ||
[[ 2010 Fall MA 26600 Holman|Back to 2010 Fall MA 26600 Holman]] | [[ 2010 Fall MA 26600 Holman|Back to 2010 Fall MA 26600 Holman]] | ||
Latest revision as of 10:53, 25 August 2010
Homework Questions/MA266Fall10
Ask and answer questions about the homework here. If asking a new question please start a new section at the bottom of the page.
I was doing supplemental problem A and found:
$ e^{-2t}(2A-2At^2+4At-3)=0. $
So if t=0 then A=3/2 And: $ e^{-2t}(-4Bt+4B-3)=0. $
So if t=0 then B=4/3
But with the variable t I wasn't sure if these answers were enough for A and B or if I needed to find more possibilities, which I had no clue how to do.
- ANSWER: You need to find values of A and B that work for all t (if there are any). Also, I think you made a mistake when taking the derivatives. Note that the derivative of $ e^{-2t} $ is $ -2 e^{-2t} $.
