Posted by david on .
dy/dx=2y^2 and if y=-1 when x=1, then when x=2, y=?
(you already posted this, but i still don't get it)
dy/dx = 2y^2
Integrating...y=2/3 y^3 + C
put 1,-1 into the equation, and solve for C
so the equation is y=(2y^3)/3-1/3
then i have to find what y equals when x=2. how do i do that. there's no x in the equation.
Now, if the problem is dx/dy = 2y^2
then x=2/3 y^3 + C and you solve.
Recheck which it is.