anonymous
  • anonymous
(d^2y/dx^2) of (dy/dx)=y^2(6-2x)
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • amistre64
you want the derivative of the derivative :)
anonymous
  • anonymous
how do i do that!
amistre64
  • amistre64
or the 3rd derivative..... hard to tell

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amistre64
  • amistre64
implicit or solve for y
amistre64
  • amistre64
implicits.....
anonymous
  • anonymous
i dont know what d^2y/dx^2 means
amistre64
  • 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
  • anonymous
oh, so i am just finding the 3rd derivative pretty much?
amistre64
  • amistre64
thats what I think..... and youll have to use implicit stuff y'=y^2(6-2x)
anonymous
  • anonymous
I forgot what implicits are):
amistre64
  • 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
  • amistre64
dy = -dx -dx y//x ; dx = 1 so.. dy = -1 -y //x
amistre64
  • 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
  • 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
  • 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
  • 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
  • anonymous
so it will be 2y y' ?
amistre64
  • amistre64
thats absolutely correct :)
amistre64
  • amistre64
after your done ; substitute the value for y' into the finished work
anonymous
  • anonymous
so if i do this problem y^2(6-x) i have to do 2y y'(6-2x)(-2)?
amistre64
  • amistre64
i find it ironic that they title me "superhero", and your named "helpless" lol
anonymous
  • anonymous
Hahaha, that is funny!
amistre64
  • amistre64
y^2(6-x) is simply the product rule with a little extra stuff jambed between the lines....
amistre64
  • amistre64
rl' + r'l is all it amounts to
amistre64
  • amistre64
r = y^2 r' = y' 2y l = (6-x) l' = -1
anonymous
  • 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
  • amistre64
squish it all together and simplify
anonymous
  • anonymous
sorry the problem was actually y^2(6-2x)
amistre64
  • amistre64
yes.... i missed the 2 from up top...but yes...thats it
amistre64
  • amistre64
you already know what y' equals, it was the original problem right?
amistre64
  • amistre64
I think they ask for y'' and give you y'=yadayada..... so this would be the final step :)
anonymous
  • anonymous
No y'=y^2(6-2x) so then 2y y'(6-2x)+y^2(-2) is y"
anonymous
  • anonymous
so i just leave it like that?
amistre64
  • amistre64
you can...but just be aware that any answers on a test might be written in the full expanded form.
anonymous
  • anonymous
how would i do that?
amistre64
  • 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
  • amistre64
2y^3(36-24x +4x^2) -2y^2
amistre64
  • amistre64
and so on and so forth...
anonymous
  • anonymous
how'd you get this part?
amistre64
  • amistre64
lets step thru this... we found y'' right? what is it?
anonymous
  • anonymous
nevermind i think i get it because you did (6-2x)(6-2x) to get (36-24x +4x^2)
amistre64
  • amistre64
yes... :)
amistre64
  • amistre64
y' was the only variable we new the "value" of, so I just replaced it and did the math
anonymous
  • 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
  • amistre64
thats right, with implicits they have to give you a way to know the x and y values.....
anonymous
  • 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
  • anonymous
*f(3)=1/4
amistre64
  • amistre64
you gotta di the inegral to find F(x) ;)
anonymous
  • anonymous
so i do the anti derivative of y'?
amistre64
  • amistre64
jsut "cross multiply" to split the variables...... yeah
amistre64
  • amistre64
dy y^2(6-2x) --- = --------- dx 1
amistre64
  • amistre64
[S] 1/(y^2) dy = [S] 6-2x dx
anonymous
  • anonymous
whats [S] ?
amistre64
  • amistre64
its my lazy version of the elongfated "S" symbol they use for the integral/anti derivative stuff
amistre64
  • amistre64
\[\int\limits_{}\]
anonymous
  • anonymous
ohhhhhh!
amistre64
  • amistre64
i loathe the "equation" editor, slows down the computer with it trying to decode all the stuff
anonymous
  • anonymous
Hahahahah, i get you!(: how did you get that from cross multiplying though?
anonymous
  • anonymous
you seperated them?
amistre64
  • 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
  • amistre64
if you havent read ahead past the "antiderivative" jargon yet... then you probably are wondering if I can do that :)
anonymous
  • anonymous
Okay so when once i get them separated, is 1/y^2 lny^2?
amistre64
  • amistre64
now now..make it easier on yourself...1/y^2 = y^-2
amistre64
  • amistre64
[S] y^-2 dy = [S] (6-2x) dx
amistre64
  • amistre64
y^-1/-1 = 6x -(2)x^2/2 + C
amistre64
  • amistre64
-1/y = -x^2 +6x + C
anonymous
  • anonymous
why is it -2x^2/2 why is it divided by two and not just x^2?
amistre64
  • 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
  • amistre64
tuen each side of..to the reciprocals.. to get "y" on top
amistre64
  • 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
  • amistre64
i took typing class and passed....honest :)
anonymous
  • anonymous
what does the // mean? sorry, i dont know how to do all this math anywhere but on paper.
amistre64
  • 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
  • amistre64
plug in your inital values and lets solve for "C" :)
anonymous
  • anonymous
so is it 1/4=(1/(x2-6x+c))
anonymous
  • anonymous
if the i.c is f(3)+.25
anonymous
  • anonymous
f(3)=.25
amistre64
  • amistre64
x^2 -6x +C = 4 x^2 -6x -4 = C right?
amistre64
  • amistre64
x = 3 and y = 1/4 yes
anonymous
  • anonymous
right.
amistre64
  • amistre64
9 - 18 - 4 = C C = -13 if I did it right :)
anonymous
  • anonymous
where is the 4 coming form?
amistre64
  • amistre64
lets see if: it works for us :) 1/4 = 1/(9 -18 -13) ...you put it there with the 1/4......right?
amistre64
  • amistre64
4 = 3^2 -6(3) + C
anonymous
  • anonymous
okay waitso the equation is 1/4=(1/(x^2-6x+c)?
amistre64
  • 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
  • anonymous
okay so when i plug in the point i get 1/4=9-18+c?
amistre64
  • 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
  • anonymous
so because 1 is on top on both sides i can cancel it?
amistre64
  • 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
  • anonymous
Okay i got it now. Now i see how you got C=13 (:
amistre64
  • amistre64
.... lol good :)
anonymous
  • anonymous
okay! Thank you so much!:D
amistre64
  • 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
  • anonymous
wait how do i find -C?
amistre64
  • amistre64
just change +C to -C and get an answer for it.....
anonymous
  • anonymous
wouldnt it just be -13?
amistre64
  • 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
  • anonymous
so i put plus and minus 13 as my answer and my problem is completely answered?(:
amistre64
  • amistre64
+c = 13 -c = --13 = 13 c = 13
anonymous
  • anonymous
okay got it! Thank you for your help!(:
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
I might be back tomorrow, hahahah.(;

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