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.
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.