Question

dy/dx=y-2y^2

Answer 1

and what do you want?

Answer 2

ode :)

Answer 3

solution equation, initial condition is (0,100) i only need a hint on finding:\[\int\limits_{}^{}1\div(y-2y ^{2})\] i dont want much more help beyond that

Answer 4

I'm not so into the integrals, sorry.

Answer 5

y' = y -2y^2 it looks like youd just reintegrate each term seperately

Answer 6

that was the starting equation, i then wrote it as dy/dx and then dy/y-2y^2=dx

Answer 7

i cant figure out how to take the integral of 1/(y-2yy)

Answer 8

helps if I keep track of the problem I spose :)

Answer 9

wiat, which equation you wanting to suit up?

Answer 10

I think you should use partial fractions

Answer 11

oh, ok
y'=y-2y^2
dy/dx=y-2y^2
dy=(y-2y^2)dx
dy/(y-2y^2)=dx
integrate both

Answer 12

ahhh partial fractions!!! that should do it! thanks!!

Answer 13

aww yeah

Answer 14

that gives me an imaginary constant of integration -.-

Answer 15

lol, did the integral work out though? I didn't really read the problem

Answer 16

would y=e^x ??

Answer 17

and by that I mean:

y = e^x + e^2x +C ?? just wondering

Answer 18

e^x - e^x^2....

Answer 19

hold on, ill upload a picture in a sec

Answer 20

can you just injtegrate the original equation (y'=y-2y^2) to (f(y)+c=1/2 y^2 - 2/3 y^3)

Answer 21

?

Answer 22

I don't see why not, and then use you initial condition for c.

Answer 23

uilfha thats pathetic, why bother doing the whole dy/dx sub when i could do that ><

Answer 24

And then I guess solve 0=1/2(100)^2-2/3(100)^3+c for c?

Answer 25

thats like -661,666.66...

Answer 26

lol, that whats confusing me. But that seems to be what they mean by the initial condition (0,100) , and thats how you would do it if you wanted the equation for x(y)