# Help me Verifying that the equation is integrable and finding its solution

How can I verify that the equation is integrable and that find its solution;

$$2y(a-x)dx+[z-y^2+(a-x)^2]dy-ydz=0$$

Honestly, I tried too much, but I got too strange results,thus I couldnt show my efforts here so sorry. Thank you for helping.

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you could add tags (differential)geometry or curves – Dylan Zhu Oct 14 '13 at 18:40
Okay tagged @DylanZhu – Nrsnr Oct 14 '13 at 18:51

Hint:

1-Verify that $$curl \left(2y(a-x),\,z-y^2+(a-x)^2,\,-y\right)=(0,0,0)$$ Actually, it is not true. May be there is a typo. I believe the second component of the vector field is $-[z-y^2+(a-x)^2]$.

2- Find the potential of the vector field $\left(2y(a-x),\,-[z-y^2+(a-x)^2],\,-y\right)$.

edit:

You need a function $f$ such that $\nabla f(x,y,z)=\left(2y(a-x),\,-[z-y^2+(a-x)^2],\,-y\right)$. Then $$f(x,y,z)=\int 2y(a-x)\,dx+C(y,z)=-y(a-x)^2+C(y,z)$$ If we derive in $y$ $$-[z-y^2+(a-x)^2]= f_y(x,y,z)=-(a-x)^2+C_y(y,z)$$ then $$C(y,z)=-zy+\frac{y^3}{3}+B(z)$$ and $$f(x,y,z)=-y(a-x)^2-zy+\frac{y^3}{3}+B(z)$$ Now, if we derive in $z$: $$-y=f_z(x,y,z)=-y+B'(z)$$ Then...

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Okay, I showed this is integrable. Now I'm trying to find its solution. For this, firstly, keep $z$ constant. And I need to find $$2y(a-x)dx+[z-y^2+(a-x)^2]dy=0$$ I tried too many times. But I cannot. Can you show me please? Thankslot for helping :) – Nrsnr Oct 14 '13 at 22:55
I used the website wolframalpha but its solution is too strange. – Nrsnr Oct 14 '13 at 22:59
I wrote some details. You are looking for a surface $f(x,y,z)=cte$. – Pocho la pantera Oct 14 '13 at 23:22
Okay thank you very much – Nrsnr Oct 14 '13 at 23:37