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anonymous
 5 years ago
dy/dx=y2y^2
anonymous
 5 years ago
dy/dx=y2y^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and what do you want?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0solution equation, initial condition is (0,100) i only need a hint on finding:\[\int\limits_{}^{}1\div(y2y ^{2})\] i dont want much more help beyond that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not so into the integrals, sorry.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y' = y 2y^2 it looks like youd just reintegrate each term seperately

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that was the starting equation, i then wrote it as dy/dx and then dy/y2y^2=dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i cant figure out how to take the integral of 1/(y2yy)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0helps if I keep track of the problem I spose :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0wiat, which equation you wanting to suit up?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you should use partial fractions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, ok y'=y2y^2 dy/dx=y2y^2 dy=(y2y^2)dx dy/(y2y^2)=dx integrate both

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh partial fractions!!! that should do it! thanks!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that gives me an imaginary constant of integration .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, did the integral work out though? I didn't really read the problem

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and by that I mean: y = e^x + e^2x +C ?? just wondering

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hold on, ill upload a picture in a sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you just injtegrate the original equation (y'=y2y^2) to (f(y)+c=1/2 y^2  2/3 y^3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't see why not, and then use you initial condition for c.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uilfha thats pathetic, why bother doing the whole dy/dx sub when i could do that ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And then I guess solve 0=1/2(100)^22/3(100)^3+c for c?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats like 661,666.66...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, 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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