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Best_Mathematician
Group Title
find general solution of this differential equation
dy/dx = y2
 one year ago
 one year ago
Best_Mathematician Group Title
find general solution of this differential equation dy/dx = y2
 one year ago
 one year ago

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iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
oh, its easily variable separable, dy/ (y2) =dx did you try integrating this ?
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
no...i got confused if i need to subtract it or what
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
no subtract. dy/ (y2) =dx got this ? can you integrate ?
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
oh ya...is the answer dw:1364766331963:dw
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
yes, now do you need to isolate 'y' also ? if not, thats your answer.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ya i need to get y out
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
as i need to find c
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
write c as ln c'
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
then use log properties.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
lets say we getta find c at x=1, y=1 how?
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
you are making this up ? general solution means we need 'c' nevertheless, y can't be 1 here.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
dw:1364766554143:dw
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ya i am making up...now lets find particular solution
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
y must be >=2 so that log term is defined...
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
no it is not
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
it is 1 and 1
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
ln of negative numbers are not defined in reals.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
then we can't find c?
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
do we need to find 'c'? and yes, if x=y=1, then c is complex number...
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
yes we need to find particular solution
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
at y(1) = 1
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
integrating gives ln (y  2) = x + c so y  2 = e^(x + c) = e^x * e*c = Ae^x y = Ae^x + 2
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
oh, good....now we don't have to actually worry about c being complex..
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
whats Ae^x
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
no we dont its calculus
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
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 ?
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
just plug in x=1, y=1 there,and you'll get A
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ok now the initial condition is y(0)=1
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
so, put x=0 y=1 this should not be any problem...
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
dw:1364767623910:dw
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
dw:1364767697343:dw
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
no, put e^c = A you can keep the general form in A and A=2 is valid.
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
you have to find a constant, not specifically 'c' A can be used, as valid constant.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
can 1 put c1 and c2 instead....i really dont know what is A
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
sure, you can even put 'Best_Mathematician' instead Best_Mathematician = 2 :P
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ya whatever lol...so A is just a constant right
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
haha thanks...do you want going further in this equation
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
this is your question...do you wanna go ?
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

iambehindyou Group TitleBest ResponseYou've already chosen the best response.1
sorry... that was latin to me :P
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ok
 one year ago
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