## anonymous 5 years ago dy/dx=y-2y^2

1. anonymous

and what do you want?

2. amistre64

ode :)

3. 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

4. anonymous

I'm not so into the integrals, sorry.

5. amistre64

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

6. anonymous

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

7. anonymous

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

8. amistre64

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

9. amistre64

wiat, which equation you wanting to suit up?

10. anonymous

I think you should use partial fractions

11. anonymous

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

12. anonymous

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

13. anonymous

aww yeah

14. anonymous

that gives me an imaginary constant of integration -.-

15. anonymous

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

16. amistre64

would y=e^x ??

17. amistre64

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

18. amistre64

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

19. anonymous

hold on, ill upload a picture in a sec

20. anonymous

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

21. anonymous

?

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

thats like -661,666.66...

26. 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)