Finding f(x,y) from partial derivatives using integration

[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

dg/dy = - (y+x²)/(x²y)

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

Mitch

 

[micromass]

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

$$-x^2y+g^\prime(y)=y-yx^2$$

 

[mitch_1211]

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?

 

[haruspex]

so g = y²/2
does that sound about right?

Well, it satisfies the original equations, right? But don't forget to allow a constant term.