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

[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.

 

[blue_raver22]

ell,

Your approach to finding the original function f(x,y) is correct. To solve for g'(x), you can integrate both sides of the differential equation you have derived. This will give you g(y) up to a constant of integration. You can then use the initial condition g(0) = 0 (assuming y = 0 is a point on the curve) to solve for the constant and find the exact function for g(y). Once you have g(y), you can substitute it back into your equation for f(x,y) to get the complete function. Good luck!