Question

Differential Equations!
Currently I have: ln y = 3 ln(x^2+4)
and it is given that y = 32 when x = 0. How do I go on from here, I'm not too sure! Fan and Medal

Answer 1

ln y= ln((x^2+4)^3)

Answer 2

Moreover, in order to apply this initial condition
y = 32 when x = 0 we need an unknown constant
I don't see one in the equation
Can you give me the initial problem statement? So I can dot all the i's and cross all the t's so to speak.

Answer 3

property of log

Answer 4

No room for an initial condition there, know what I mean?

Answer 5

I'm not so sure what you mean Miracrown...

Answer 6

y=(x^2+4)^3

Answer 7

thats what we got here

Answer 8

I have
ln y = 3 ln(x^2+4) + ln(32/64)

Answer 9

Hm, give me the initial DE and I'll show you

Answer 10

that is you got after solving

Answer 11

initially what was the Q

Answer 12

Sure, but I never learnt anything about dotting stuff. It is question 3 here
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w12_qp_33.pdf

Answer 13

Question 4 I mean

Answer 14

that's a logarithmic equation

Answer 15

Ah, okay, looks like you made good progress

Answer 16

which Q??

Answer 17

After taking the logs out I get y = 1/2(x^2+4)^3
but the answers say
y = 1/2 (x^2+4)
So apparently there's no cubed...

Answer 18

Quetsion 4

Answer 19

wait

Answer 20

but the right hand side doesn't need a logarithmic anti-derivative

Answer 21

\[\frac{ dy }{ y }=\frac{ 6x*dx }{ x^2+4 }\]

Answer 22

now we need to integrate it

Answer 23

Yeah got that which is:
ln y = 3 ln(x^2+4)

Answer 24

ln y =3*ln(x^2+4) + c

Answer 25

no no we also add a integration constant

Answer 26

c

Answer 27

Oh yeah forgot about that :/
Then we put in the values for y = 32 and x = 0 right?

Answer 28

ln y =3*ln(x^2+4) + c
now in this put y=32 and x=0

Answer 29

and calculate c

Answer 30

c = ln (32/64)

Answer 31

ln 32=3*ln(0+4)+c
ln32=3*ln4 +c
ln32=ln4^3 + c
ln32=ln64 +c
ln 32-ln 64=c
c=ln(32/64)=ln(1/2)=ln 1 - ln 2 =0- ln 2

Answer 32

yeahhhhh

Answer 33

So what do we do next? We've got
ln y = 3 ln (x^2+4) + ln(32/64)

Answer 34

yepppppppppppppppppp

Answer 35

lny=ln(x^2+4)^3+ln (1/2)
32/64=1/2 and ln(1/2)=ln 1 - ln 2 =-ln 2 as ln 1 =0
ln y =(ln (x^2+4)^3-ln 2
ln y = ln[{(x^2+4)^3}/2]
now ln on oth side will get cancelled

Answer 36

so the final answer is
y = 1/2 (x^2+4)^3?