Did I separate the variables correctly for these two questions?: Attachment 30552 If not, what did I do wrong? Which option would work?

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- March 28th 2014, 01:12 PMcanyouhelpSeparating Variables to find an equation?
Did I separate the variables correctly for these two questions?: Attachment 30552 If not, what did I do wrong? Which option would work?

- March 29th 2014, 08:02 PMHallsofIvyRe: Separating Variables to find an equation?
Six is wrong, seven is correct. Again, it is impossible to say what you did wrong when you don't show what you did!

- March 30th 2014, 07:14 AMcanyouhelpRe: Separating Variables to find an equation?
Okay, so the first thing I did for number 6 was factor the equation..because the question says to do that. Then I seperated x and y. Y on the left and x on the right. Then I found the antiderivative. Or at least I thought I did. Maybe that's where I messed up? Also, I wasn't sure what to do with y(2)=10. I'm not used to using initial conditions, and our professor didn't explain them.

- March 30th 2014, 08:05 AMromsekRe: Separating Variables to find an equation?
don't post in English. Post in math. Learn LaTex or post clear images of your work.

- March 30th 2014, 03:14 PMcanyouhelpRe: Separating Variables to find an equation?
Sorry I'm not sure how to use LaTex. So for now I'll just post my work the way I can: For #6, I just retried it and got a completely different answer than before: first I factored -4y+36 to get 4(9-y) and -4(y-9). So then I multiplied both sides by (y-9). And after separating them more I get ln(y-9)=-4x+C So I plug in (2,10) and get something that would make no sense at all. C would have to equal 40. What did I do wrong? Obviously something went way wrong.

- March 30th 2014, 08:27 PMromsekRe: Separating Variables to find an equation?
At the point that you are doing differential equations your algebra needs to be second nature and yours is far from it. I really don't know what you ended up doing but somehow you ended up with the correct answer up to $\ln\left|y-9\right| = -4x + C$

Now $y(2)=10$ so

$\ln(10-9)=-4(2)+C$

$\ln(1)=-8+C$

$0 = -8 + C$

$C = 8$

$\ln\left|y-9\right|=-4x + 8$

i.e. choice 1

you really need to get your algebra nailed down. - March 31st 2014, 06:55 AMcanyouhelpRe: Separating Variables to find an equation?