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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

1 Answer 1

up vote 1 down vote accepted

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

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