1. ## Gradient system

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

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

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 $\displaystyle x'=f(x,y)$ and $\displaystyle y'=g(x,y).$ Then $\displaystyle x'=-\nabla V$ implies $\displaystyle f(x,y)=-\frac{\partial V}{\partial x}$ and $\displaystyle g(x,y)=-\frac{\partial V}{\partial y}$.

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