Find the particular solution

$$\large xy^2 dx + e^x dy = 0$$ when $x \rightarrow \infty$ and $y \rightarrow \frac 12$

what do those approach mean -_-

Question

Find the particular solution

$$\large xy^2 dx + e^x dy = 0$$ when $x \rightarrow \infty$ and $y \rightarrow \frac 12$

what do those approach mean -_-

anonymous · Answer

too hard, I quit.

apoorvk · Answer

Now divide the whole equation by dx

whaddaya get? (no, am not latexing this for you ^_^)

lgbasallote · Answer

i have the integral thingies already...i need to know what those approaching means...for subbing

anonymous · Answer

$$
\large xy^2 dx + e^x dy = 0\
\large xy^2 dx =-e^x dy \

\large-\frac {dy}{y^2}= x e^{-x} dx\
\large \frac 1 y = e^{-x}( 1-x) + C\
\large   y =\frac 1{ e^{-x}( 1-x) + C}\
C=2,\,  \text { from the condition given }
$$

anonymous · Answer

I'm pretty sure, the right answer, is the right answer :s :3

lgbasallote · Answer

shouldnt there be a negative somewhere sir @eliassaab o.O

anonymous · Answer

The antidrevative of -1/y^2 is 1/y