Question

Hard problem.

If y = 2x/1-x^2

prove that:

(1+x^2)y'' = 2(2y -x)y'

Answer 1

oh wow this one is a little tricky give me a minuite for this one

Answer 2

@Taylor0402 , I am not joking. Try your luck.

Answer 3

umm...wow this is really hard i cant solve this im only in 8th grade but this should help

Answer 4

@dan815

Answer 5

xD

Answer 6

just plug and chug

Answer 7

why no sub?

Answer 8

wait so u cant solve for y'' or y' ??

Answer 9

how can u solve it if u dont make use of y=f(X)

Answer 10

http://prntscr.com/94nioe

Answer 11

No wolfram alpha in exams to simplify.

Answer 12

just simplify it, its only algebra

Answer 13

How about reaching the proof from the given ?

Answer 14

just work with that expression, i dont wanna write it alll out, just start simplfying it

Answer 15

(1-x^2)y = 2x

-2xy + (1-x^2)y' = 2

-2y -2xy' + (-2x)y' + (1-x^2)y'' = 0

(1-x^2)y'' = 2(y - 2x)y'

Can't lol

Answer 16

no the expression from wolfram man

Answer 17

i found waht y' is and y'' and subbed it in and got that huge expression

Answer 18

y''=-(4 x (3+x^2))/(-1+x^2)^3,
y'=(2 (1+x^2))/(-1+x^2)^2,
y = 2x/(1-x^2)

sub into
(1-x^2)y'' = 2(y - 2x)y'

and simplify just algebra then

Answer 19

You mean (1+x^2)y'' = 2(2y -x)y

I tried to prove it and got that
(1-x^2)y = 2x

-2xy + (1-x^2)y' = 2

-2y -2xy' + (-2x)y' + (1-x^2)y'' = 0

(1-x^2)y'' = 2(y - 2x)y'

It has a solution in 3 step but it's not for human begins.

Answer 20

no u gotta subbb

Answer 21

this is kinda 3 steps

Answer 22

u find y' then y''
sub and thenull see its the same thing

Answer 23

(1+x^2)((4 x (3+x^2))/(-1+x^2)^3) = 2(2( 2x/(1-x^2)) -x)(2 (1+x^2))/(-1+x^2)^2)

Answer 24

I need a year to simplify that xD

Answer 25

@dan815 , any flexible solution ?

Answer 26

what do u mean lol xD u can already see its kidn of true looking at it

Answer 27

LOL

Answer 28

(1+x^2)((4 x (3+x^2))/(-1+x^2)^3) = 2(2( 2x/(1-x^2)) -x)(2 (1+x^2))/(-1+x^2)^2)

Ok, If I told you evaluate this expression , what would you say other than huh ?

Answer 29

xDD noo its simple man

Answer 30

okay start by writing this expression as one fraction http://prntscr.com/94nu0p

Answer 31

come on u can doo thisss grade 5 stuff gogogogogoggo

Answer 32

Ok , see this easy solution that no one would think of :

y = 2x/1-x^2

y = 2 tan(tan^-1(x))/1-tan^2(tan^-1(x)) = tan(2tan^-1(x))

y' = sec^2(2arctan(x)) * 2 * 1/(1+x^2)

y' = (1+ tan^2(2arctan(x)) * 2/(1+x^2)

(1+x^2)y' = 2(1+y^2)

(1+x^2)y'' + 2xy' = 4yy'

(1+x^2)y'' = 2(2y - x)y'

Surely I can, but I am lazy to do all that O.O

Answer 33

@dan815 , still wanna me calculate the expression ?

Answer 34

yes

Answer 35

-4x(1+x^2)(3+x^2)/ (-1+ x^2)^3

Answer 36

=.= okay can i go now lol xD this simplification u can literally do in ur head lol

Answer 37

(8x/ 1-x^2 - 2x ) 2(1+x^2)/(-1+x^2)^2

16x(1+x^2)/(-1+x^2)^@ - 4x(1+x^2)/(-1+x^2)^2

Answer 38

@ = 2

Answer 39

we have already a common denominator

Answer 40

12x(1+x^2)/(-1+x^2)^2

Answer 41

That's the L.H.S

Answer 42

-4x(1+x^2)(3+x^2)/ (-1+ x^2)^3

so
12x(-1+x^2)/(-1+x^2)^3

now we have to prove that -4x(3+x^2) = 12x

Answer 43

http://prntscr.com/94o16i

Answer 44

u shud tget that

Answer 45

do it on paper its not neat here

Answer 46

Thanks, Go help other ppl, I won't simplify that.

Answer 47

why :S

Answer 48

i dont understand why u are being lazy...

Answer 49

it takes 1 min litereally