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find general solution of this differential equation dy/dx = y-2

Calculus1
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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 ?

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Other answers:

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
you didn't get this ? : ln (y - 2) = x + c so y - 2 = e^(x + c) = e^x * e*c = Ae^x y = Ae^x + 2 what you have doubt with ?
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 ?
ya lol..... Use a technological tool and Euler's method with ten steps of size 0.1 to estimate y(1). Make certain to include supporting screen captures.
sorry... that was latin to me :P
ok

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