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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. | ||
| − | + | ---- | |
I was doing supplemental problem A and found: | I was doing supplemental problem A and found: | ||
| − | e^ | + | |
| − | + | <math>e^{-2t}(2A-2At^2+4At-3)=0.</math> | |
| + | |||
| + | So if t=0 then A=3/2 | ||
And: | And: | ||
| − | e^-2t(-4Bt+4B-3)=0 | + | <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 here | ||
[[ 2010 Fall MA 26600 Holman|Back to 2010 Fall MA 26600 Holman]] | [[ 2010 Fall MA 26600 Holman|Back to 2010 Fall MA 26600 Holman]] | ||
Revision as of 08:59, 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 here
