i would greatly apperciate it if someone were to take a look at my work and tell where im wrong or right thx :)

oh yeah i'll probably be bugging ya guys all night

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- Oct 11th 2006, 05:12 PMcyberdx16someone wanna look over my work?
i would greatly apperciate it if someone were to take a look at my work and tell where im wrong or right thx :)

oh yeah i'll probably be bugging ya guys all night - Oct 11th 2006, 05:58 PMtopsquark
First problem: Just a typo. Should be dx/dt = 3y^2/(6y). You had "x^2."

Third problem:

You have y' = 4/5 + (3/5)y'

So (5/5)y' - (3/5)y' = 4/5

(2/5)y' = 4/5

You've got an extra negative sign.

Fourth problem:

x^2 + y^3 = 5y

Take the first derivative:

2x + 3y^2*y' = 5 y'

Take the second derivative:

2 + 6y*y' + 3y^2*y'' = 5y''

3y^2*y'' - 5y'' = -2 - 6y*y'

y'' = -(6y*y' + 2)/(3y^2 - 5)

You've got the point (x,y) from the problem and y' at this point from the last problem.

And for the record, you can TRY to bug me all night, but I'm going to be in bed by 11. :D

-Dan - Oct 11th 2006, 06:48 PMcyberdx16
how did u get (the bold part?) y"=2+6y*y'

**+3y^2***y"=5y" - Oct 11th 2006, 07:04 PMThePerfectHacker
- Oct 11th 2006, 07:48 PMcyberdx16
so does anyone know if i did the 2nd problem right?

- Oct 11th 2006, 08:54 PMCaptainBlack

The final answer is right, but your notation could be confusing, better to

write it as:

dy/dx=(3 x^2)/(x^3+1).

The argument which leads to your solution however is confused. It should

go more like:

y=f(g(x)), where g(x)=x^3+1, and f'(x)=1/x.

Then by the chain rule:

dy/dx = f'(g(x)) g'(x) = [1/(x^3+1)] 3x^2 =(3x^2)/(x^3+1).

RonL