Question

Given that y=0 when x=1, solve the differential equation \[xy {dy \over dx}=y^2 + 4\] obtaining an expression for y^2 in terms of x.

~~ Sorry for posting it again, but the last one was already quite full and long, so I think it's better to start a-fresh. I didn't understand the last time... ~ So, could you go step by step? :D thanks!

Answer 1

where you able to get to that point

Answer 2

Yes, I was able to get to that point. Very explicit, and very nice :) So, from here?

Answer 3

Its in reverse order

Answer 4

First question: How did you get \[\int\limits {y \over y^2 +4}dy ---->{\ln(y^2+4) \over 2}=lnx?\]

Answer 5

hope this explains it. if not let me know

Answer 6

It explains the 1/x for me... but where does the y go (the top y) in the first integral?

Answer 7

The y goes to heaven.

Answer 8

if we had ln(y^2). When we take the derivative of that function it will be y/y^2.

Answer 9

So the y on top was taken care of by remembering that the give equation is in the form of (ln(ax^n))'

Answer 10

And the 2 came from? --

Answer 11

look at the last image i uploaded. Do you see the n on the top

Answer 12

yes

Answer 13

since the derivative of (ln(y^2 + 4)) is equal to (2y/(y^2 + 4)). where the 2 is my n. Since in our equation our n was not available, i had to divide my result by my n which is 2.

Answer 14

ok

Answer 15

And how does lnx^2 become x^2 + c?

Answer 16

It because when you take the anti derivative of a given function, there are infinite number of functions that can equal to your result. so the convention is to just put a constant in this case c at the end of your result.

Answer 17

OK

Answer 18

does that make sense

Answer 19

I understand the rest. Thank you so much!!! :D I'm doomed in my real exam though.. I'd never think of getting the answer this way, and then being correct throughout the way... :/

Answer 20

I believe you can do it. Just don't CRAM. Learn the general form of these derivatives and anti derivatives and also if you can prove these general forms, i guarantee you that it will become part of your brain. : )

Answer 21

Haha... One month to go :/ Not even completed the syllabus yet... Ahh! Anyway, thanks for the help and support! I really appreciate it!

Answer 22

No problem, let me know if you need any help. Also use http://www.khanacademy.org he has lots of tutorials that might help you. Good Luck.