(d^2y/dx^2) of (dy/dx)=y^2(6-2x)

- anonymous

(d^2y/dx^2) of (dy/dx)=y^2(6-2x)

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- amistre64

you want the derivative of the derivative :)

- anonymous

how do i do that!

- amistre64

or the 3rd derivative..... hard to tell

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## More answers

- amistre64

implicit or solve for y

- amistre64

implicits.....

- anonymous

i dont know what d^2y/dx^2 means

- amistre64

d^2y/dx^2 is just the 2nd derivative of an equation, for example:
y = 2x^2
dy/dx = 4x
d2y/dx2 = 4

- anonymous

oh, so i am just finding the 3rd derivative pretty much?

- amistre64

thats what I think..... and youll have to use implicit stuff
y'=y^2(6-2x)

- anonymous

I forgot what implicits are):

- amistre64

implicits are the same steps as normal derivatives except you "keep" the implicits in the equation to factor out in the end...like this:
x+xy = 4
d(x) + d(xy) = d(4)
dx + dx y + x dy = 0
solve for dy

- amistre64

dy = -dx -dx y//x ; dx = 1 so..
dy = -1 -y //x

- amistre64

y' = y^2(6-2x)
y'' = y' 2y(6-2x) + y^2(-2x)x' ; x' = 1
y'' = y' 12y -4xy - 2xy^2 is what I get

- anonymous

how do i find the derivative of y, like x's you bring the exponent to the front?
like if it says y^2 what do i do?

- amistre64

watch what we do with derivatives :)
d (2x^3) dx (6x^2)
---- = ----
dx dx
now dx/dx = 1 usually so we omit it.... right?

- amistre64

d (y^2) dy (2y)
---- = ----
dx dx
dy/dx IS y' or the first derivative. it doesnt cancel so we just keep it in the equation as y'

- anonymous

so it will be 2y y' ?

- amistre64

thats absolutely correct :)

- amistre64

after your done ; substitute the value for y' into the finished work

- anonymous

so if i do this problem y^2(6-x) i have to do 2y y'(6-2x)(-2)?

- amistre64

i find it ironic that they title me "superhero", and your named "helpless" lol

- anonymous

Hahaha, that is funny!

- amistre64

y^2(6-x) is simply the product rule with a little extra stuff jambed between the lines....

- amistre64

rl' + r'l is all it amounts to

- amistre64

r = y^2 r' = y' 2y
l = (6-x) l' = -1

- anonymous

oh! so it'll be 2y y'(6-2x)+y^2(-2)? or am i going off on the wrong track with the y's again?

- amistre64

squish it all together and simplify

- anonymous

sorry the problem was actually y^2(6-2x)

- amistre64

yes.... i missed the 2 from up top...but yes...thats it

- amistre64

you already know what y' equals, it was the original problem right?

- amistre64

I think they ask for y'' and give you y'=yadayada..... so this would be the final step :)

- anonymous

No y'=y^2(6-2x) so then 2y y'(6-2x)+y^2(-2) is y"

- anonymous

so i just leave it like that?

- amistre64

you can...but just be aware that any answers on a test might be written in the full expanded form.

- anonymous

how would i do that?

- amistre64

just replace the y' with y^2(6-2x) and work the math :)
2y (y^2)(6-2x)(6-2x)+y^2(-2)

- amistre64

2y^3(36-24x +4x^2) -2y^2

- amistre64

and so on and so forth...

- anonymous

how'd you get this part?

- amistre64

lets step thru this...
we found y'' right? what is it?

- anonymous

nevermind i think i get it because you did (6-2x)(6-2x) to get (36-24x +4x^2)

- amistre64

yes... :)

- amistre64

y' was the only variable we new the "value" of, so I just replaced it and did the math

- anonymous

'& that's the answer right?
So if it gives me a point, i would just plug the values in for my x and y? then solve for that?

- amistre64

thats right, with implicits they have to give you a way to know the x and y values.....

- anonymous

okay i think i got it. But part b says to find y=f(x) by solving (dy/dx)=y^2(6-2x) using the initial condition f(3)+1/4

- anonymous

*f(3)=1/4

- amistre64

you gotta di the inegral to find F(x) ;)

- anonymous

so i do the anti derivative of y'?

- amistre64

jsut "cross multiply" to split the variables......
yeah

- amistre64

dy y^2(6-2x)
--- = ---------
dx 1

- amistre64

[S] 1/(y^2) dy = [S] 6-2x dx

- anonymous

whats [S] ?

- amistre64

its my lazy version of the elongfated "S" symbol they use for the integral/anti derivative stuff

- amistre64

\[\int\limits_{}\]

- anonymous

ohhhhhh!

- amistre64

i loathe the "equation" editor, slows down the computer with it trying to decode all the stuff

- anonymous

Hahahahah, i get you!(: how did you get that from cross multiplying though?

- anonymous

you seperated them?

- amistre64

this is our original problem right?
dy y^2(6-2x)
--- = ---------
dx 1
we cross multiply to get the "y" stuff to one side and the "x" stuff to the other so we can inegrate them separately

- amistre64

if you havent read ahead past the "antiderivative" jargon yet... then you probably are wondering if I can do that :)

- anonymous

Okay so when once i get them separated, is 1/y^2 lny^2?

- amistre64

now now..make it easier on yourself...1/y^2 = y^-2

- amistre64

[S] y^-2 dy = [S] (6-2x) dx

- amistre64

y^-1/-1 = 6x -(2)x^2/2 + C

- amistre64

-1/y = -x^2 +6x + C

- anonymous

why is it -2x^2/2 why is it divided by two and not just x^2?

- amistre64

.... its a place holder so I dont lose track of it and make more mistakes than ususal :)
-2x becomes -2x^2/2 simplify .... -x^2

- amistre64

tuen each side of..to the reciprocals.. to get "y" on top

- amistre64

-y = 1 // -x^2 +6x + C get rid of the (-) by dividing both sides by -1
y = 1 // x^2 -6x - C
and we should have it

- amistre64

i took typing class and passed....honest :)

- anonymous

what does the // mean? sorry, i dont know how to do all this math anywhere but on paper.

- amistre64

I use that to mean the there is a larger fraction bar that separates the top and bottom, it just means that everything before it is on top and everything after it is on bottom. otherwise id look like this:
1/x^2 -6x -C which can be misleading....

- amistre64

plug in your inital values and lets solve for "C" :)

- anonymous

so is it 1/4=(1/(x2-6x+c))

- anonymous

if the i.c is f(3)+.25

- anonymous

f(3)=.25

- amistre64

x^2 -6x +C = 4
x^2 -6x -4 = C right?

- amistre64

x = 3 and y = 1/4 yes

- anonymous

right.

- amistre64

9 - 18 - 4 = C C = -13 if I did it right :)

- anonymous

where is the 4 coming form?

- amistre64

lets see if: it works for us :)
1/4 = 1/(9 -18 -13)
...you put it there with the 1/4......right?

- amistre64

4 = 3^2 -6(3) + C

- anonymous

okay waitso the equation is 1/4=(1/(x^2-6x+c)?

- amistre64

yep, cross multiply or jsut flip them both over or jsut notice that the bottom parts need to match up and equal each other

- anonymous

okay so when i plug in the point i get 1/4=9-18+c?

- amistre64

no...lets start at the beginning of this:
1
y = -----------
x^2 -6x +C
f(3) = .25 = 1/4
1 1
-- = --------------
4 (3)^2 -6(3) +C
4 = 9 -18 + C

- anonymous

so because 1 is on top on both sides i can cancel it?

- amistre64

....not "because" the 1 is on top..... but yeah. we are wanting to make that bottom right hand side the same number as the bottom of the left hand side.
We can either cross multiply and get them both on top.
We can flip them over to their reciprocal counterparts (which is what I did).
Or, we can simply notice that the only thing that matters in this equation is that the bottom numbers have to equal and we can ignore anything else :)

- anonymous

Okay i got it now. Now i see how you got C=13 (:

- amistre64

.... lol good :)

- anonymous

okay! Thank you so much!:D

- amistre64

did we solve our problem? because I am wondering if that +C changes if we are looking for -C.... better get 2 answers so you can double check yourself

- anonymous

wait how do i find -C?

- amistre64

just change +C to -C and get an answer for it.....

- anonymous

wouldnt it just be -13?

- amistre64

dunno, my brains not giving me an answer lol
4 = 9 -18 - C
C = 9-4-18
C = 5 - 18 = -13
were good.....

- anonymous

so i put plus and minus 13 as my answer and my problem is completely answered?(:

- amistre64

+c = 13
-c = --13 = 13
c = 13

- anonymous

okay got it! Thank you for your help!(:

- amistre64

youre welcome :)

- anonymous

I might be back tomorrow, hahahah.(;

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