Question

This is for integration properties of constants.
I have lnx=ln(1+u^2)/2)+c
when I take e to the both sides I get x=e^c (1+u^2)/2 the initial value problem is y(-1)=1 and the U becomes y/x I need to solve it for C which is the question how do I get C to the other side and solve it.

Answer 1

The given constraint is y(-1) = 1... which means when x=-1, y=1... :-)

Answer 2

You have to solve it for the value of C because of integration. The question is e^(c) what are the rules about moving the C around in e functions is e^c the same thing as c*e

Answer 3

i suggest you find 'c' before taking e raised to both sides...

Answer 4

I moved my equation around and I get c= x/sqrt(x^2+y^2/x^2))
I just am wondering what is the rules about constants. like if c*e/x^-1/2 is the same as e^c(x(1/2)

Answer 5

why is e^c the same thing as c*e

Answer 6

it isn't..

Answer 7

Hey again... e^c and c^e are not same...
And what I mean by *The given constraint is y(-1) = 1... which means when x=-1, y=1..*
is that substitute x=-1, y=1, then you will get numerical value of 'c'...

Hope this makes sense...

Answer 8

I did do the substitution I just got confused on how I obtained c from the
eqation lnx=ln(1+(y/x)^2)/2 + c
or the lnx= sqrt(1+(y/x)^2)+c
I get C= -1/sqrt(2) when I solve with initial values

Answer 9

i would do it this way,
x=e^c (1+u^2)/2
when x=-1, u= -1
so, e^c =-1
i think, you also got this ??
so,
x=e^c (1+u^2)/2
--->x=-(1+u^2)/2

Answer 10

no need to find c at all, use e^c=-1

Answer 11

Thats what I did and I get e^c = -1/sqrt(2)
so then i just put it into x=e^c(sqrt u^2+1) thats my answer ?

Answer 12

also, c actually would be imaginary,
e^c=-1
c= ln (-1) = i \(\pi\)

Answer 13

how u got get e^c = -1/sqrt(2) ?

Answer 14

This is where I'm getting confused... the same thing I got but it would give no real value of c...
The only thing we can do from there is , use e^(ipi) = -1, that means taking log to -1 ...
Otherwise if real c is asked, no such possible value...

B/w I agree with hartnn... not getting the value you found...

Answer 15

e^c x/sqrt of x^2+y^2/x^2 this value is what (y^2/x^2)+1 =

then it is -1 =x and 1 = y so -1/sqrt((1+1/1)) -1/sqrt(2) the values of are all positive because of it being squared.

Answer 16

and it is e^c=x/sqrt ((x^2+y^2)/x^2)

Answer 17

ohh, its square root, then its correct...

Answer 18

Thanks for your help again I think you helped me last night also. I got my seperation problem done is the ln(ln(u) anything I just left it as x=c-lnln(u)

Answer 19

welcome :) yes.
is your 2nd statement , a new problem ? i didn't get what u said...

Answer 20

I asked you last night to about seperation of x(dy/dx)= yln(xy)
I got it seperated in terms of v and x so dv/dx=-1lnv
then integrated them to solve and i got lnln(v)=-x+c

Answer 21

ohh! yes, now i remember.... actually i didn't solve it fully, so got confused , where did ln ln v came from...

Answer 22

seperate the two gets dv/lnv =-1dx
dv/u is ln(u) u=lnv so you get lnln(v)

Answer 23

as far as i remember, u (or v) was xy
and not in terms of ln....i hope you solved that correctly.

Answer 24

dv/dx -v/x= (v/x)ln v
this is variable separable, if you try it. is what you put
but it simplifies into dv/dx= -1ln(v) v=xy and y=v/x

Answer 25

ok,
dv/dx = (v/x)(1+ln v)
v(1+ln v) dv = dx/x
now integrating (by parts for left, and ln x for right.)

Answer 26

dv/dx= -1ln(v) is apparantely, not correct....

Answer 27

uh oh I always forget h

Answer 28

how to do stuff properly

Answer 29

umm..practicing....