nfornick

Can you try this? (x+3)/(x^2-3) as x -> infinity

Re: Mobius a*b*c<1 but not a*b*c<0 "at least ONE of them must be a fraction" means one or two may be a fraction "one of them will be a fraction (say a) and as it is on the bottom line it will be inverted and will be large. It will also be larger then the other two numbers added together (b + c)" a*b*c<1 does not imply 1/a>b+c say a=1/3, b=1, c=2

Show that a/c + b/a + c/b > a+b+c, for a, b, c > 0 and a*b*c<1

my point is my H(x,y) is not the solution and I want to ask where did I made mistake

insight of ElijahJones's simplication i solved it using integrating factor, but this seems not my question's aim

this is the exe under the title exact diff eqt so i just follow what i've learn

Solve e^(xy)(ycosx-sinx)+xe^(xy)cosxdy/dx=0 I've checked this is exact and found that H(x,y)=e^(xy)(y^2cosy-cosy)/(1+y^2)+y^2/(1+y^2)C(y) As the answer is simply e^(xy)cosx=C, I guess I made a mistake before differentiating H(x,y) to get C(y) What's wrong with my H(x,y)?