Question

Differential equations help!
y*dx + (3+3x-y)*dy = 0
The topic is integrating factors
The answer is 4xy^3 + 4y^3 - y^4 = C

Answer 1

hi bahrom

Answer 2

want me to use that list thingy

Answer 3

u dont have to

Answer 4

just a solution, its integrating factors, its just my brains are dead

Answer 5

what do you mean by integrating factors?

Answer 6

you know when you let something be [e^Integral(f(x;y))dx] then multiply across by it to get a derivative of some expression on the right and some ez integral on the left or vice versa

Answer 7

oh

Answer 8

so you have y + dy/dx ( 3 + 3x -y) = 0

Answer 9

well yeah then move y over

Answer 10

does it mean you're basically trying to prove the answer?

Answer 11

no I just know the answer

Answer 12

well for integrating factors we want dy/dx + p(x)y = g(x)

Answer 13

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

Answer 14

oh okay then i can solve it i think

Answer 15

integrating factor is e^x, lemme try this will let u guys know how it goes

Answer 16

wait no its not, but at least now i know the expression, tryin to solve afk

Answer 17

here, this is called exact equation
http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx

Answer 18

M(x,y) + N(x,y)dy/dx = 0

Answer 19

k

Answer 20

dont kill yourself bahrom
this website is buggy, everytime i type something it freezes

Answer 21

still there?

Answer 22

have you done partial derivatives?

Answer 23

i have, but as i said my brains are not functionin at all, just too tired

Answer 24

we are looking for a function Fx (x,y) + Fy (x,y) dy/dx = 0, where Fx is the partial derivative of F(x,y) with respect to x (treat y as a constant) and Fy is the partial derivative of F(x,y) with respect to y (treat x as a constant)

Answer 25

ok .... :(

Answer 26

We have the equation M(x,y) dx + N(x,y) dy = 0 or
M (x,y) + N(x,y)dy/dx = 0

M(x,y) = y and N(x,y) = 3 + 3x - y
remember we can solve this and its called "exact" if M(x,y) = Fx (x,y) and N(x,y) = Fy (x,y), where Fx is the partial derivative of F(x,y) with respect to x (treat y as a constant) and Fy is the partial derivative of F(x,y) with respect to y (treat x as a constant) so if we have Fx (x,Y) + Fy (x,y) dy/dx = 0, then F(x,y) is our solution .

ok so the last thing , assuming there exists such a F(x,y) it must be the case that Fxy = Fyx which is true for all continuous F(x,y). By substituting Fx = M and Fy = N we have
My = Nx .
So here Mx = 0 , and Ny = -1 . hmmm, not exact
ok now we move to plan B and introduce an integrating factor

http://www.cliffsnotes.com/study_guide/Integrating-Factors.topicArticleId-19736,articleId-19711.html

Answer 27

here is another website http://www.sosmath.com/diffeq/first/exact/exact.html
and they have a non exact first order ode