Question

what is the general solution of (xy^3)dx + (e^(x^2))dy = 0

my progress:
(xy^3)dx = -(e^(x^2))dy
xdx/(e^(x^2)) = dy/y^3
......
i know how to integrate dy/y^3 but does anyone know how to integrate xdx/(e^(x^2)) ?

Answer 1

let x^2 = t
see if this helps

Answer 2

uh...lemme try

Answer 3

It's quite easy if you just get the hang of how to do it.

Answer 4

wait, does that mean -dx/(e^(x^2)) becomes -t/(e^(t^2)) ?

Answer 5

nop, try again

Answer 6

No, keep trying.

Answer 7

so if x^2 = t then -dx/(e^(x^2)) becomes -dx/(e^t) right? then I integrated via ln therefore
-dx/(e^t) becomes lne^t then
ln(e^(x^2))?

Answer 8

uhm...is it right?

Answer 9

nop
it becomes
dt/2e^t

please recheck

Answer 10

since the integration of (e^(x^2)) [where e^u du = e^u + C] is 2(e^(x^2)) right? i was kinda thinking of that as well but what I was wondering was how to integrate them fully. because 2(e^(x^2) ) is not the derivative of dx [meaning dx/x = ln|x| +C ] so what should I do with (e^(x^2)) so it can be a derivative of dx?

Answer 11

wait, I think i'm getting it.
since -dx/(e^(x^2)) becomes dt/2e^t then it becomes (-1/2)(dt/e^t) right?
then:
(-1/2)(e^-t)(dt) = (1/2)(e^-t) + C right?

Answer 12

integration of (e^(x^2)) [where e^u du = e^u + C] is NOT 2(e^(x^2))..

and yes, latter thing you said was right

Answer 13

thank you! (oh and (e^(x^2)) = 2x(e^(x^2)) right? sorry. my bad ^^;;)

Answer 14

the derivative of
\[ d\ e^{-x^2} = e^{-x^2} d\ (-x^2) = -e^{x^2} 2x \ dx\]