# Easy Equation, stuck on the constant.

• January 4th 2011, 04:28 AM
craig
Easy Equation, stuck on the constant.
Let $\omega = (3y^2 + 2y)dx + (6xy + 2x + y^2)dy$.

Find potential function $f$ for $\omega$.

We know $\frac{\partial f}{\partial x} = 3y^2 + 2y$, so $f = 3xy^2 + 2xy + g(y)$

Differentiating wrt $y$, $\frac{\partial f}{\partial y} = 6xy + 2x + g'(y) = 6xy + 2x + y^2$.

So we have $g(x) = \frac{1}{3}y^3$ + constant.

The answers then simply state that the constant is just zero, not quite sure how they get to this?

• January 4th 2011, 04:59 AM
Sudharaka
Quote:

Originally Posted by craig
Let $\omega = (3y^2 + 2y)dx + (6xy + 2x + y^2)dy$.

Find potential function $f$ for $\omega$.

We know $\frac{\partial f}{\partial x} = 3y^2 + 2y$, so $f = 3xy^2 + 2xy + g(x)$

Differentiating wrt $y$, $\frac{\partial f}{\partial y} = 6xy + 2x + g'(x) = 6xy + 2x + y^2$.

So we have $g(x) = \frac{2}{3}y^3$ + constant.

The answers then simply state that the constant is just zero, not quite sure how they get to this?

Dear craig,

You have a arbitary constant. Therefore you can take any value you like for the constant. For simplicity they may have taken it to by equal to zero.
• January 4th 2011, 05:00 AM
craig
Ahh thankyou. I was debating whether you could basically choose your value of the constant.

• January 4th 2011, 11:46 AM
General
That g(x) should be g(y).
• January 4th 2011, 12:42 PM
dwsmith
Quote:

Originally Posted by craig
Let $\omega = (3y^2 + 2y)dx + (6xy + 2x + y^2)dy$.

Find potential function $f$ for $\omega$.

We know $\frac{\partial f}{\partial x} = 3y^2 + 2y$, so $f = 3xy^2 + 2xy + g(x)$

Differentiating wrt $y$, $\frac{\partial f}{\partial y} = 6xy + 2x + g'(x) = 6xy + 2x + y^2$.

So we have $g(x) = \frac{2}{3}y^3$ + constant.

The answers then simply state that the constant is just zero, not quite sure how they get to this?

You set your $\displaystyle\frac{df}{dy}=N(x,y)$

Then you have

$\displaystyle 6xy+2x+g'(y)=6xy+2x+y^2\Rightarrow g'(y)=y^2$

$\displaystyle \int g'(y)dy=\int y^2dy\Rightarrow g(y)=\frac{y^3}{3}+c$

I have a different solution for your g(y) solution then what you obtained. I don't think you should have a 2/3 just a 1/3.
• January 4th 2011, 03:55 PM
craig
Thanks for the last two posts. You're both right, typos due to typing far too quickly lol.

Cheers for spotting those :)