Find the solution to 3y^2y'+16x=2xy^3, which is limited when x-+infinite

Question

Find the solution to 3y^2y'+16x=2xy^3, which is limited when x->+infinite

anonymous · Answer

i came to y^3=C*e^(x^2)-8

anonymous · Answer

apply variable separable method, that should get us to the answer

anonymous · Answer

$$y^3= 8+k*e^{3*(x^2)}$$

anonymous · Answer

and what does the part about limits, mean?
as in, y tends to some value, right?
what value?

anonymous · Answer

it only says that i have to find solution of this equation, which is limited when x goes to $$\infty$$

anonymous · Answer

and the result should be y=2

anonymous · Answer

well then, if y is supposed to be limited, the only way that might happen is if the term containing x vanishes as x -> +infinity, right?
the only way that might happen is if k=0 and then we might get y=2, if that is the answer.

anonymous · Answer

oh, y=2 is the answer :P