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Mar19-11, 12:28 PM
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Quote Quote by LCKurtz View Post
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: Qx or [itex]Q_x[/itex].
Quote Quote by pat666 View Post
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]


ps should be 4x to the power of 3/2 , doesn't seem to like my tex.
Quote Quote by LCKurtz View Post
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 Quote by pat666 View Post
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]
[/tex] = 13/3? because it =s 7??

am I doing something wrong with my checking or is my solution wrong?

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 = 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.