Given that a system is a gradient system, here's how to find it's potential function V. Suppose that $x'=f(x,y)$ and $y'=g(x,y).$ Then $x'=-\nabla V$ implies $f(x,y)=-\frac{\partial V}{\partial x}$ and $g(x,y)=-\frac{\partial V}{\partial y}$.

Use this procedure to find V for the following gradient system:
$x'=y^2+y cos x$
$y'=2xy+sin x$

2. Originally Posted by splash
Given that a system is a gradient system, here's how to find it's potential function V. Suppose that $x'=f(x,y)$ and $y'=g(x,y).$ Then $x'=-\nabla V$ implies $f(x,y)=-\frac{\partial V}{\partial x}$ and $g(x,y)=-\frac{\partial V}{\partial y}$.

Use this procedure to find V for the following gradient system:
$x'=y^2+y cos x$
$y'=2xy+sin x$
Can you state your question differently.
You want to find a vector field $F(x,y)$ such as what?
$\nabla F=$
In that case,
$F_x=y^2+y\cos x$
$F_y=2xy+\sin x$
Thus, (in first equation)
$F=xy^2+y\sin x+f(y)$
Where $f(y)$ is to be determined.
But then,
$F_y=2xy+\sin x+f'(y)$
So, $f'(y)=0$
So, $f(y)=C$
Thus, it seems the function,
$F(x,y)=xy^2+y\sin x+C$