Quote by LCKurtz
Don't quit yet, you almost have it. Take x > 0 to dispense with the  signs:
[tex]R(x) = e^{\int\frac 1{2x}dx}= e^{\frac 1 2 \int\frac 1 x}\, dx
= e^{\frac 1 2 \ln x} = e^{\ln(x^{\frac 1 2})} = x^{\frac 1 2}[/tex]
Also, a TeX note, you can use the partial symbol
[tex]\partial Q/\partial x[/tex]
but subscripts are even easier: Q_{x} or [itex]Q_x[/itex].

Quote by pat666
Thanks LCKurtz I now have a general and particular solution, would you mind checking my results:
[tex] u(x,y)=4x^{3/2}+2*\sqrt(x)*y^2+y=c [/tex]
[tex] c=13/3 [/tex]
[tex] 13/3=4x^{3/2}+2*\sqrt(x)*y^2+y [/tex]
THANKS
ps should be 4x to the power of 3/2 , doesn't seem to like my tex.

Quote by LCKurtz
You should be able to check that yourself. Check that du = Pdx + Qdy with the exact form and that (x,y)=(1,1) works. If not, try to fix it.

Quote by pat666
I don't know what you mean by Check that du = Pdx + Qdy, du as in the differential of
[tex] 4x^{3/2}+2*\sqrt(x)*y^2+y [/tex]
also should [tex]
4*1^{3/2}+2*\sqrt(1)*1^2+1
[/tex] = 13/3? because it =s 7??
am I doing something wrong with my checking or is my solution wrong?
Thanks

After you multiplied by the integrating factor x
^{1/2} you should have gotten this DE, which you didn't show:
[tex](*)\ (2x^{\frac 1 2}+ 2x^{\frac 1 2}y^2)dx+4x^{\frac 1 2}y dy = 0[/tex]
Presumably you have checked that this equation is exact and you have a proposed solution u(x,y) written above. The way to check whether your solution is correct is to calculate du = u
_{x}dx + u
_{y}dy and see if it agrees with the exact equation (*) above.
Once you have that you need to check that x = 1, y = 1 works in your final answer.