dy/dx=y-2y^2

- anonymous

dy/dx=y-2y^2

- jamiebookeater

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- anonymous

and what do you want?

- amistre64

ode :)

- anonymous

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

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## More answers

- anonymous

I'm not so into the integrals, sorry.

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

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

- amistre64

wiat, which equation you wanting to suit up?

- anonymous

I think you should use partial fractions

- anonymous

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

- anonymous

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

- anonymous

aww yeah

- anonymous

that gives me an imaginary constant of integration -.-

- anonymous

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

- amistre64

would y=e^x ??

- amistre64

and by that I mean:
y = e^x + e^2x +C ?? just wondering

- amistre64

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

- anonymous

hold on, ill upload a picture in a sec

- anonymous

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

- anonymous

?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

thats like -661,666.66...

- anonymous

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)

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