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9709/33/o/n/12 please find an expression for y in terms of x ?

the variables c and y are related by the differential equation (x^2+4)dy/dx=6xy
it is given that y= 32 when x=0 find an expression for y in terms of x

after differentiation and substitution ln(y)=3ln(x^2+4)-ln(2)
now the expression for y in terms of x is supposed to be y=1/2x^2+2 but i get a different answer.....

so any help please ?
Reply 1
Avatar for davros
davros
16
Original post by missworld96
the variables c and y are related by the differential equation (x^2+4)dy/dx=6xy
it is given that y= 32 when x=0 find an expression for y in terms of x

after differentiation and substitution ln(y)=3ln(x^2+4)-ln(2)
now the expression for y in terms of x is supposed to be y=1/2x^2+2 but i get a different answer.....

so any help please ?


I get

y=12(x2+4)3\displaystyle y = \dfrac{1}{2}(x^2 + 4)^3 if you've written the question correctly.

Where does this come from?
Original post by davros
I get

y=12(x2+4)3\displaystyle y = \dfrac{1}{2}(x^2 + 4)^3 if you've written the question correctly.

Where does this come from?


Ah-ha! I know this one.

There is a reference in the thread title.

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w12_qp_33.pdf
Reply 3
Avatar for davros
davros
16
Original post by Mr M


If you recognize the question then that's really sad :smile:

I thought the title was a reference number, but I wouldn't have known where to check!

Do you agree with my answer (not fully concentrating as I'm watching the repeat of Tough Young Teachers at the moment :biggrin: )?
Original post by davros
...


First hit in Google.

I'm watching TYT too (dull as dishwasher).

Your answer is right.
(edited 10 years ago)
Reply 5
Avatar for davros
davros
16
Original post by Mr M
First hit in Google.

I'm watching TYT too (dull as dishwasher).

Your answer is right.


I like watching posh kids get humiliated :smile:

I'll watch the repeat of Bad Education later as a bit of an antidote.


Thanks!

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