## Finding a function given its partial derivatives, stuck on finding g'(x)

Hi all,

I have the following partial derivatives

∂f/∂x = cos(x)sin(x)-xy2

∂f/∂y = y - yx2

I need to find the original function, f(x,y).

I know that df = (∂f/∂x)dx + (∂f/∂y)dy

and hence

f(x,y) = ∫∂f/∂x dx + g(y) = -1/2(x2y2+cos2(x)) + g(y)

Then to find g(y) I took the partial derivative of f(x,y) that I just found wrt y and equate to the original given ∂f/∂y

i.e -x2y +g'(y) = y-x2
which is a first order differential equation of the form

dg/dy = - (y+x2)/(x2y)

and now I am not sure how to proceed and solve this DE

Mitch
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 Quote by mitch_1211 Hi all, I have the following partial derivatives ∂f/∂x = cos(x)sin(x)-xy2 ∂f/∂y = y - yx2 I need to find the original function, f(x,y). I know that df = (∂f/∂x)dx + (∂f/∂y)dy and hence f(x,y) = ∫∂f/∂x dx + g(y) = -1/2(x2y2+cos2(x)) + g(y) Then to find g(y) I took the partial derivative of f(x,y) that I just found wrt y and equate to the original given ∂f/∂y i.e -x2y +g'(y) = y-x2 which is a first order differential equation of the form
That's a typo, no? It should be

$$-x^2y+g^\prime(y)=y-yx^2$$
 oh you're right, I must have typed it out incorrectly from my notes. so it becomes dg/dy = y ∴ ∫dg = ∫y dy so g = y2/2 does that sound about right?

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