Question

help me solve this DE please. again. thank you!

(x^3 + xy^2 -1)dx + x(x^2 + y^2 +1)dy = 0

Answer 1

\[(x^3 + xy^2 -1) \ \text{d}x + x(x^2 + y^2 +1) \ \text{d}y = 0\]

Answer 2

is this a kind of obvious substitution?

Answer 3

\[M\text{d}x+N\text{d}y=0\]i think we must work on exact diff equations ha?

Answer 4

i can think that the substitution : x=cos t , y=sin t willl greatly simplify this...have to work out though...

Answer 5

ok. exact. i'll try.

how to tell if what method am i going to use ?

Answer 6

i'm stuck.

Answer 7

\[M=x^3+xy^2−1\]\[N=x^3+xy^2+x\]\[\frac{\partial M}{\partial y}=2xy\]\[\frac{\partial N}{\partial x}=3x^2+y^2+1\]...i'm stuck too

Answer 8

@mukushla ,can u give some reference material for this type.....i have done such problems few years ago,then i can also try....

Answer 9

http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx

Answer 10

sorry, i don't have any corrections here. but how do you do it with that kind of substitution?

Answer 11

let
t=x^3+xy^2
then try it.

Answer 12

substitution is difficult if i'm going to use obvious substitution. :(

Answer 13

@sami-21 i tried it but im stuck on that too
@hartnn we're not allowed to do sub like that

Answer 14

sami is it working?

Answer 15

let me try.

Answer 16

not working with that substitution :(

Answer 17

u let y= sin t ,x = cos t

so the answer is x^2+y^2=1

Answer 18

and we dont need to solve it further

Answer 19

ohh,silly mistake....if i make x = cos t ,then i cannot tell that y will be sin t....okkk

Answer 20

@lilMissMindset

is this from exact equations part?

Answer 21

i really don't know. it's an exercise. and i don't know what to do with this.

Answer 22

is this for real?

Answer 23

i can think of equation like this\[(x^3 + xy^2 -y) \ \text{d}x + (yx^2 + y^3 +x) \ \text{d}y = 0\]but i dont know what to do with\[(x^3 + xy^2 -1) \ \text{d}x + x(x^2 + y^2 +1) \ \text{d}y = 0\]

Answer 24

maybe @Traxter or @UnkleRhaukus can help us

Answer 25

I don't think the equation is exact so you can't use that method.
Will have a look at it in a minute.