find a function$ f$ such that $\nabla f = F$

Question

Determine whether or not $F$ is a conservative vector field. If it is, find a funcion $f$ such that $F=\nabla f$.$$F(x,y)=y^2e^{xy}\vec i+(1+xy)e^{xy}\vec j$$

I tried following the method in my book. I integrated the coefficient of i with respect to x to get $y^4(e^xy + y^4g'(y))$ and that lead to a really messy value of $g'(y)$. I don't think its supposed to be that messy, so I must be doing something wrong, but I can't figure out what. Can someone please show me the complete process here?

Answer 1

Welcome to exact differential equations.

I'll write this in the form $P \ dx + Q \ dy$. Notice that

$$ \frac{\partial P}{\partial y} = 2ye^{xy} + xy^2 e^{xy} $$

and

$$ \frac{\partial Q}{\partial x} = 2ye^{xy} + xy^2 e^{xy} $$

(check this!) so certainly this is conservative. Now we need a potential function. Do

$$ \int P \ dx = \int y^2 e^{xy} \ dx = ye^{xy} + g(y) $$

Now differentiate this in $y$ and set it equal to the other bit, $Q$.

$$ xye^{xy} + e^{xy} + g'(y) = (1 + xy)e^{xy} $$

This leaves $g'(y) = 0$, so $g(y) = C$, done.

See here.

Answer 2

Integrating the coefficient of $\vec i$ wrt $x$ should have given $ye^{xy}+g(y)$. I am not sure how you got to your expression.

Differentiating this with respect to $y$ gives $e^{xy}(1+yx)+g'(y)$. Comparing with coefficient of $\vec j$ gives that $g(y)=constant$. So you have found your $f$.

Answer 3

If you integrate $y^2e^{xy}$ with respect to $x$ you should get $ye^{xy}+g(y)$ for some function $g(y)$. The $y$-derivative of this has to be equal to $(1+xy)e^{xy}$...