Finding the integrating factor

  • Thread starter newtomath
  • Start date
  • #1
37
0
I am having trouble with the below:

[ 4* (x^3/y^2) + (3/y)] dx + [3*(x/y^2) +4y]dy=0

I found My= -8x^3y^-3 - 3y^-2 and Nx= 3y^-2
i then subtracted Nx from My and divided by [3*(x/y^2) +4y]


[-8x^3y^-3 - 6y^-2] / [3*(x/y^2) +4y]. can you guys give me a hint as to where my error is?
 

Answers and Replies

  • #2
478
32
So I multiplied the original equation by y^2 on both sides and got:

(4x^3+3y) dx + (3x+4y^3) dy = 0

and this equation is exact. You have to remember that I(x) = exp^(Integral[(M_y-N_x)/N,x]). In this case, since M_y=N_x, you get exp(Integral[0,x]) = 1.
 
  • #3
37
0
Thanks. I figured that (Nx-My)/M did the trick
 

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