Question

x^2 + 2xy + (x^2 - 1/3x^3 - x^2y)dy/dx = 0
Solve this DE by finding a function of the form u = u(y), as an integrating factor.

Answer 1

oh, okay, so what is a function u = u(y) such that both u and u' are present in the given equation

Answer 2

x^2 and 1/3x^3 ?

Answer 3

u = ?
u' = ?

Answer 4

u = 1/3 u ^3
u' = x^2 ?

Answer 5

not u^3, and u' needs to be the first derivative of u

Answer 6

sorry, i meant x!

Answer 7

or would it be y?

Answer 8

oh, okay
no, though; try to replace as much of the given equation as possible by u and u'

Answer 9

ok and then from there ..

Answer 10

did you get something for u and u'?

Answer 11

what do you mean did i get something for them? i thought you said just to sub them into the formula
which gives me

u' + 2xy + (u' - u + u'y)dy/dx = 0

Answer 12

no, i said:
oh, okay, so what is a function u = u(y) such that both u and u' are present in the given equation

Answer 13

like...

Answer 14

don't just substitute u and u' for x and y

Answer 15

u is a function and u' is the derivative of that function

Answer 16

i didn't? i subbed
u = 1/3 x ^3
u' = x^2

Answer 17

okay that's right - so sorry! - now, add more to u and u' so that you can reduce t he equation further by substitution

Answer 18

you said try to replace as much as the given equatoin as possible with u and u' ..

Answer 19

yep

Answer 20

like x^2*y and 2xy

Answer 21

i'm supposed to add more to them? it's kind of unclear.

Answer 22

yeah sorry

Answer 23

actually, start with
-(x^2 + 2xy)dx = (x^2 - 1/3x^3 - x^2y)dy
:-)

Answer 24

ok.. so then dont start with the integrating factor subsitution?