Question

Find an integrating factor and solve...
See attachment.

Answer 1

I know that the equation isn't exact, so you need to multiply something in order to have it exact. The problem lies in what to do afterwards...integrate in respect to x...possible integrate in respect to y,

Answer 2

I tried integrating in respect to x and take the derivative in terms of y, but I got 6x^2y^2 when there's x^2y^3. If I integrate in terms of x and y I do get the exponents but I am left with a number.

Answer 3

I know that for exact equations...at least for the My part you need to integrate with respect to x + h(y) and then take the derivative in terms of y which produces whatever it is + h'(y) and then set it equal to the Nx which should have something cancel out, but that's not the case here. I mean it works but the exponents are wrong.

Answer 4

alright I found an example that might help me out...I'll come back here if I run into trouble

Answer 5

arghhhhh ran into trouble

Answer 6

@myko

Answer 7

how is this even possible? I mean I did all the steps correctly...everything cancels out, yet the answer is supposed to be x^2y^3 = c

Answer 8

unless I multiply the 4/3 with the exponents... but is that legal?

Answer 9

is the integrating factor -x^2y^2

Answer 10

O_O from what I read off of this site. http://www.sosmath.com/diffeq/first/intfactor/Example/Example.html

if it's not exact you need to do dm/dy - dn/dx divided by N

Answer 11

I got -2xy/3x^2y which becomes -2/3x and then I integrate that... it becomes