Question

Determine whether the DE is exact. IF it is exact, solve it. If not, find an appropriate integrating factor and solve the resulting equation.
[\cosx dx+(1+2/y)sinx dy=0]\
I got -ln(cosx) +y+2lny=c using an integrating factor of 1/sinx. Can anyone confirm this?

Answer 1

it's not exact...

Answer 2

Is the integrating factor 1/x?

Answer 3

sorry was away.. let me check real quick.. im also askin for help here..

Answer 4

Yeah I saw the linear algebra question. Wish I could help. Maybe in a couple months...

Answer 5

Integrating factor is either (Nx - My)/M = ((1+2/y)-0)/Cosx.. well apparently it's not since we don't have anything in terms of one variable..

Answer 6

Let's try:
(My-Nx)/N oh and also by the way e to all of that...

Answer 7

wait a second.. i think i was wrong..

Answer 8

Shouldn't Nx be cosx +2/y cosx?

Answer 9

sorry now that i wrote it on paper.. yea..

Answer 10

I made the same mistake, you have to multiply it out

Answer 11

so yea u have:
(Nx-My)/M = (Cosx + (2/y) Cosx)/Cosx = 1 + 2/y

Answer 12

Your integrating factor is:
e^(Integral of ( 1+2/y )dy) = y^2e^y

Answer 13

Now check to see if it's exact..
My = Cosx(y^2e^y + e^y*2y) = ye^yCosx(y+2)

Answer 14

Nx = y^2e^yCosx+2ye^yCosx = ye^yCosx(y+2)
My=Nx => this is exact=>y^2e^y is an integrating factor..

Answer 15

What did you get as the answer? I used p(x) instead of p(y) but it should come out the same...