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 Quote by LCKurtz Don't quit yet, you almost have it. Take x > 0 to dispense with the || signs: $$R(x) = e^{\int-\frac 1{2x}dx}= e^{-\frac 1 2 \int\frac 1 x}\, dx = e^{-\frac 1 2 \ln x} = e^{\ln(x^{-\frac 1 2})} = x^{-\frac 1 2}$$ Also, a TeX note, you can use the partial symbol $$\partial Q/\partial x$$ but subscripts are even easier: Qx or $Q_x$.
 Quote by pat666 Thanks LCKurtz I now have a general and particular solution, would you mind checking my results: $$u(x,y)=4x^{3/2}+2*\sqrt(x)*y^2+y=c$$ $$c=13/3$$ $$13/3=4x^{3/2}+2*\sqrt(x)*y^2+y$$ 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 $$4x^{3/2}+2*\sqrt(x)*y^2+y$$ also should $$4*1^{3/2}+2*\sqrt(1)*1^2+1$$ = 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:

$$(*)\ (2x^{\frac 1 2}+ 2x^{-\frac 1 2}y^2)dx+4x^{\frac 1 2}y dy = 0$$

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 = uxdx + uydy 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.