How Can Clairaut's Theorem Help Solve for Function g(y) in Partial Derivatives?

[matpo39]

hi, I am having some trouble with this problem:

A certain function f(x,y) is known to have partial derivatives of the form

(partial)f/(partial)x = 2ycos(2x)+y^3*x^2+g(y)
(partial)f/(partial)y= sin(2x)+x^3*y^2+4x+1

Please note that g is a function of y only. Use the equality of mixed partial derivatives (Clairaut's Theorem) to find the function g up to an arbitrary additive constant. then find all the functions f.

i was able to attempt the first part and i got

g'(y)= 4
and then intagrating that i find that g(y)= 4y+c_1

i then inteagrated the two partial functions and i got

f(x)= ysin(2x)+(1/3)y^3*x^3+4yx+c_2
f(y)= ysin(2x)+(1/3)y^3*x^3+4yx+y+c_3

i have no idea on what i should do next, if anyone can help me out, that would be great.

thanks

 

[diegojco]

basically are saying you that the function f(x,y) has a gradient equal to (∂f/∂x,∂f/∂y). The condition for a field to be gradient of some function f is the equality of mixed partial derivatives.

 

[wais]

Hi there,

It looks like you have made some progress on this problem already, which is great! Let's take a look at your work and see if we can help you move forward.

First, you are correct in using Clairaut's Theorem to find the function g. This theorem states that if the partial derivatives of a function are continuous, then the order in which you take the partial derivatives does not matter. In other words, the mixed partial derivatives are equal. So, we can set the two partial derivatives equal to each other and solve for g(y):

2ycos(2x)+y^3*x^2+g(y) = sin(2x)+x^3*y^2+4x+1

Since g is a function of y only, we can treat x as a constant. This means that the terms with x in them will cancel out, leaving us with:

g(y) = 4x+1

Now, we can integrate both sides with respect to y to find g(y):

∫ g(y) dy = ∫ (4x+1) dy

g(y) = 4xy+y+c

Note that we have added a constant of integration, c, since we are integrating with respect to y.

Next, we can integrate the two partial functions with respect to x and y, respectively. However, we need to be careful with the integration limits since we are dealing with a function of two variables. Let's start with the first function:

∫ (∂f/∂x) dx = ∫ (2ycos(2x)+y^3*x^2+g(y)) dx

f(x,y) = ysin(2x)+(1/3)y^3*x^3+g(y)x+c_1

Note that we have added a constant of integration, c_1, since we are integrating with respect to x. Now, let's integrate the second function:

∫ (∂f/∂y) dy = ∫ (sin(2x)+x^3*y^2+4x+1) dy

f(x,y) = ysin(2x)+(1/3)y^3*x^3+4yx+g(y)y+c_2

Note that we have added a constant of integration, c_2, since we are integrating with respect to y. However, we also have an extra term, g(y)y