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oh, its easily variable separable,
dy/ (y-2) =dx
did you try integrating this ?

no...i got confused if i need to subtract it or what

no subtract.
dy/ (y-2) =dx
got this ? can you integrate ?

oh ya...is the answer |dw:1364766331963:dw|

yes, now do you need to isolate 'y' also ?
if not, thats your answer.

ya i need to get y out

as i need to find c

write c as ln c'

then use log properties.

lets say we getta find c at x=1, y=1 how?

you are making this up ?
general solution means we need 'c'
nevertheless, y can't be 1 here.

|dw:1364766554143:dw|

ya i am making up...now lets find particular solution

y must be >=2 so that log term is defined...

no it is not

it is 1 and 1

ln of negative numbers are not defined in reals.

then we can't find c?

do we need to find 'c'?
and yes, if x=y=1, then c is complex number...

yes we need to find particular solution

at y(1) = 1

integrating gives
ln (y - 2) = x + c
so y - 2 = e^(x + c) = e^x * e*c = Ae^x
y = Ae^x + 2

oh, good....now we don't have to actually worry about c being complex..

whats Ae^x

no we dont its calculus

just plug in x=1, y=1 there,and you'll get A

ok now the initial condition is y(0)=1

so, put x=0 y=1
this should not be any problem...

|dw:1364767623910:dw|

|dw:1364767697343:dw|

no, put e^c = A
you can keep the general form in A
and A=-2 is valid.

you have to find a constant, not specifically 'c'
A can be used, as valid constant.

can 1 put c1 and c2 instead....i really dont know what is A

sure, you can even put 'Best_Mathematician' instead
Best_Mathematician = -2 :P

ya whatever lol...so A is just a constant right

yup.

haha thanks...do you want going further in this equation

this is your question...do you wanna go ?

sorry... that was latin to me :P

ok