Question

\[(x+xy^2)dx + e^{x^2}ydy =0\]

Answer 1

\[(x+xy^2)dx + e^{x^2}ydy=0\]\[x(1+y^2)dx =- e^{x^2}ydy\]\[-\frac{x}{e^{x^2}}dx=\frac{y}{1+y^2}dy\]\[-\frac{1}{2}e^{-x^2}d(x^2)=\frac{1}{2(1+y^2)}d(y^2)\]\[\frac{1}{2}e^{-x^2}+C=\frac{1}{2}\ln |1+y^2|\]\[e^{-x^2}+C=\ln |1+y^2|\]Does it look fine?

Answer 2

it looks correct to me.

Answer 3

i dont see anything wrong here

Answer 4

Oh!! Thanks!!
Should I further simplify it and make y as the subject or just leave it here?

Answer 5

no, point....
that will just complicate the equation.
exponential of exponential.....

Answer 6

Okay, thanks! :)

Answer 7

welcome ^_^