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

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

Mathematics
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you want the derivative of the derivative :)
how do i do that!
or the 3rd derivative..... hard to tell

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Other answers:

implicit or solve for y
implicits.....
i dont know what d^2y/dx^2 means
d^2y/dx^2 is just the 2nd derivative of an equation, for example: y = 2x^2 dy/dx = 4x d2y/dx2 = 4
oh, so i am just finding the 3rd derivative pretty much?
thats what I think..... and youll have to use implicit stuff y'=y^2(6-2x)
I forgot what implicits are):
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
dy = -dx -dx y//x ; dx = 1 so.. dy = -1 -y //x
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
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?
watch what we do with derivatives :) d (2x^3) dx (6x^2) ---- = ---- dx dx now dx/dx = 1 usually so we omit it.... right?
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'
so it will be 2y y' ?
thats absolutely correct :)
after your done ; substitute the value for y' into the finished work
so if i do this problem y^2(6-x) i have to do 2y y'(6-2x)(-2)?
i find it ironic that they title me "superhero", and your named "helpless" lol
Hahaha, that is funny!
y^2(6-x) is simply the product rule with a little extra stuff jambed between the lines....
rl' + r'l is all it amounts to
r = y^2 r' = y' 2y l = (6-x) l' = -1
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?
squish it all together and simplify
sorry the problem was actually y^2(6-2x)
yes.... i missed the 2 from up top...but yes...thats it
you already know what y' equals, it was the original problem right?
I think they ask for y'' and give you y'=yadayada..... so this would be the final step :)
No y'=y^2(6-2x) so then 2y y'(6-2x)+y^2(-2) is y"
so i just leave it like that?
you can...but just be aware that any answers on a test might be written in the full expanded form.
how would i do that?
just replace the y' with y^2(6-2x) and work the math :) 2y (y^2)(6-2x)(6-2x)+y^2(-2)
2y^3(36-24x +4x^2) -2y^2
and so on and so forth...
how'd you get this part?
lets step thru this... we found y'' right? what is it?
nevermind i think i get it because you did (6-2x)(6-2x) to get (36-24x +4x^2)
yes... :)
y' was the only variable we new the "value" of, so I just replaced it and did the math
'& 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?
thats right, with implicits they have to give you a way to know the x and y values.....
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
*f(3)=1/4
you gotta di the inegral to find F(x) ;)
so i do the anti derivative of y'?
jsut "cross multiply" to split the variables...... yeah
dy y^2(6-2x) --- = --------- dx 1
[S] 1/(y^2) dy = [S] 6-2x dx
whats [S] ?
its my lazy version of the elongfated "S" symbol they use for the integral/anti derivative stuff
\[\int\limits_{}\]
ohhhhhh!
i loathe the "equation" editor, slows down the computer with it trying to decode all the stuff
Hahahahah, i get you!(: how did you get that from cross multiplying though?
you seperated them?
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
if you havent read ahead past the "antiderivative" jargon yet... then you probably are wondering if I can do that :)
Okay so when once i get them separated, is 1/y^2 lny^2?
now now..make it easier on yourself...1/y^2 = y^-2
[S] y^-2 dy = [S] (6-2x) dx
y^-1/-1 = 6x -(2)x^2/2 + C
-1/y = -x^2 +6x + C
why is it -2x^2/2 why is it divided by two and not just x^2?
.... 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
tuen each side of..to the reciprocals.. to get "y" on top
-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
i took typing class and passed....honest :)
what does the // mean? sorry, i dont know how to do all this math anywhere but on paper.
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....
plug in your inital values and lets solve for "C" :)
so is it 1/4=(1/(x2-6x+c))
if the i.c is f(3)+.25
f(3)=.25
x^2 -6x +C = 4 x^2 -6x -4 = C right?
x = 3 and y = 1/4 yes
right.
9 - 18 - 4 = C C = -13 if I did it right :)
where is the 4 coming form?
lets see if: it works for us :) 1/4 = 1/(9 -18 -13) ...you put it there with the 1/4......right?
4 = 3^2 -6(3) + C
okay waitso the equation is 1/4=(1/(x^2-6x+c)?
yep, cross multiply or jsut flip them both over or jsut notice that the bottom parts need to match up and equal each other
okay so when i plug in the point i get 1/4=9-18+c?
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
so because 1 is on top on both sides i can cancel it?
....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 :)
Okay i got it now. Now i see how you got C=13 (:
.... lol good :)
okay! Thank you so much!:D
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
wait how do i find -C?
just change +C to -C and get an answer for it.....
wouldnt it just be -13?
dunno, my brains not giving me an answer lol 4 = 9 -18 - C C = 9-4-18 C = 5 - 18 = -13 were good.....
so i put plus and minus 13 as my answer and my problem is completely answered?(:
+c = 13 -c = --13 = 13 c = 13
okay got it! Thank you for your help!(:
youre welcome :)
I might be back tomorrow, hahahah.(;

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