Solve for Function M in Exact Equation | 65 Characters

[Math10]

Homework Statement

Find the function M such that M(x, y)dx+(x^2-y^2)dy=0 is exact.

Homework Equations

This is an exact equation.

The Attempt at a Solution

Here's my work:

integral of (x^2-y^2)dy=yx^2-y^3/3+h(x)=C
2xy+h'(x)=M(x, y)
But the answer in the book is M(x, y)=2xy+f(x). How do I get to the answer from there?

 

[zoki85]

By putting h'(x)=f(x)

 

[gopher_p]

Homework Statement

Find the function M such that M(x, y)dx+(x^2-y^2)dy=0 is exact.

Homework Equations

This is an exact equation.

The Attempt at a Solution

Here's my work:

integral of (x^2-y^2)dy=yx^2-y^3/3+h(x)=C
2xy+h'(x)=M(x, y)
But the answer in the book is M(x, y)=2xy+f(x). How do I get to the answer from there?

So you have $M(x,y)=2xy+h'(x)$ and the book has $M(x,y)=2xy+f(x)$. Seems to me that your answer is basically the same as the book's; you just have different names for the the "extra" function of $x$.

 

[Mark44]

Homework Statement

Find the function M such that M(x, y)dx+(x^2-y^2)dy=0 is exact.

Homework Equations

This is an exact equation.

The Attempt at a Solution

Here's my work:

integral of (x^2-y^2)dy=yx^2-y^3/3+h(x)=C
2xy+h'(x)=M(x, y)
But the answer in the book is M(x, y)=2xy+f(x). How do I get to the answer from there?

They want a function M such that M_y = N_x.
IOW,
$$\frac{\partial M}{\partial y} = \frac{\partial (x^2 - y^2)}{\partial x} = 2x$$
If you integrate 2x with respect to y, you get 2xy + some arbitrary function of x. Taking the partial with respect to y of a function that involves only x is similar to taking the derivative of a constant.

 

[HallsofIvy]

Let f(x,y) be any function of two variables and assume that x and y are themselves functions of some variable, t. Then we could write f as a function of the single variable t and, by the chain rule, df/dt= (\partial f/\partial x) dx/dt+ (\partial f/\partial y)dy/dt. In terms of "differentials" that is df= (\partial f/\partial x)dx+ (\partial f/\partial y)dy.

An expression of the form M(x,y)dx+ N(x,y)dy is an "exact differential" if and only if there exist a function f such that \partial f/\partial x= M(x, y) and \partial f/\partial y= N(x, y). Of course, if that is true, we have \partial M/\partial y= \partial^2 f/\partial x\partial y and \partial N/\partial x= \partial^2 f/\partial y\partial x. But those "mixed second partial derivatives" must be equal so to be an "exact differential" we must have \partial M/\partial y= \partial N/\partial x

Here, N= x^2- y^2 so that \partial N/\partial x= 2x. We must have \partial M/\partial y= 2x so that, integrating with respect to y while holding x constant, we have M= 2xy+ f(x). "f(x)" is the constant of integration- since we integrate with respect to y while holding x constant, that "constant of integration" can be any function of x.

 

[Math10]

They want a function M such that M_y = N_x.
IOW,
$$\frac{\partial M}{\partial y} = \frac{\partial (x^2 - y^2)}{\partial x} = 2x$$
If you integrate 2x with respect to y, you get 2xy + some arbitrary function of x. Taking the partial with respect to y of a function that involves only x is similar to taking the derivative of a constant.

Thank you so much, @Mark44!