Question

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

Answer 1

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

Answer 2

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

Answer 3

\[y^3= 8+k*e^{3*(x^2)}\]

Answer 4

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

Answer 5

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

Answer 6

and the result should be y=2

Answer 7

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.

Answer 8

oh, y=2 is the answer :P