Expert: Socrates - 9/28/2005

Question
hi,

I'm trying to solve the DE:
cos(x)dx + (1 + (2/y))sin(x)dy = 0
using an intergration factor

I let M = cos(x) and N= (1 + (2/y))sin(x)
then,
  My = 0, and Nx = cos(x)(1 + (2/y))

so, (Nx - My) / M = (1 + (2/y))

I found the integrating factor to be exp(y) * y^2
then I multiplied the DE by the integrating factor and obtained

(exp(y)*y^2)cos(x)dx + (exp(y)*y^2)sin(x)dy = 0


Here is where I get lost, I can't get My = Nx, so that the equation is exact.


I really appreciate any advice you can offer, thank you,
    Carrie

Answer
First, the integrating factor really is (e^y)(y^2), as you found correctly. But, did you multiply the DE through by (e^y)(y^2) correctly? It appears you forgot the factor (1+2/y) from the original DE.

Multiplying the DE by (e^y)(y^2) gives

(e^y)(y^2)cos(x)dx + (e^y)(y^2)(1 + (2/y))sin(x)dy = 0
 
(e^y)(y^2)cos(x)dx + (e^y)(y^2 + 2y)sin(x)dy = 0

Now, for the new M and N for this last equation,

dM/dy = (e^y)(y^2 + 2y)cos(x)

dN/dx = (e^y)(y^2 + 2y)cos(x)

So the equation is now exact.

Integrating (e^y)(y^2)cos(x) with respect to x gives
(e^y)(y^2)sin(x)

So the solution to the DE is given implicitly by
(e^y)(y^2)sin(x) = c


Note: if you can solve this DE by any method you like, I would prefer to simply divide the DE by sin(x) and get

cot(x)dx + (1+2/y)dy = 0

this equation is separable, and integrating both sides of

(1+2/y)dy = -cot(x)dx

gives

y + 2 ln(y) = -ln(sin(x)) + c

applying exp to both sides, we get

(e^y)(y^2) = c/sin(x)

so

(e^y)(y^2)sin(x) = c ,

and this agrees with the answer found above.

This just seems easier to me.

Good Luck,
Paul