# Separating Variables to find an equation?

• Mar 28th 2014, 12:12 PM
canyouhelp
Separating 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?
• Mar 29th 2014, 07:02 PM
HallsofIvy
Re: 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!
• Mar 30th 2014, 06:14 AM
canyouhelp
Re: Separating Variables to find an equation?
Quote:

Originally Posted by HallsofIvy
Six is wrong, seven is correct. Again, it is impossible to say what you did wrong when you don't show what you did!

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.
• Mar 30th 2014, 07:05 AM
romsek
Re: Separating Variables to find an equation?
don't post in English. Post in math. Learn LaTex or post clear images of your work.
• Mar 30th 2014, 02:14 PM
canyouhelp
Re: Separating Variables to find an equation?
Quote:

Originally Posted by romsek
don't post in English. Post in math. Learn LaTex or post clear images of your work.

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.
• Mar 30th 2014, 07:27 PM
romsek
Re: Separating Variables to find an equation?
Quote:

Originally Posted by canyouhelp
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.

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.
• Mar 31st 2014, 05:55 AM
canyouhelp
Re: Separating Variables to find an equation?
Quote:

Originally Posted by romsek
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.

Thank you! I think the problem I had was when I wrote it out after getting ln(y-9)=-4x+C is that I made the 4 positive by accident and it messed up the entire thing.