Find g(x)/h(y) for a given F(x,y)

[songoku]

Homework Statement

Given that $F(x,y)=f(2x^2 +5y^2)$ and satisfies $h(y) \frac{\partial F}{\partial x}+g(x) \frac{\partial F}{\partial y}=0$, find $\frac{g(x)}{h(y)}$

Relevant Equations

Partial derivative

$$F(x,y)=f(2x^2+5y^2)$$

$$\frac{\partial F}{\partial x}=f'(2x^2+5y^2) . (4x)$$

$$\frac{\partial F}{\partial y}=f'(2x^2+5y^2).(10y)$$

$f'(2x^2+5y^2)=\frac{\partial F}{\partial x} . \frac{1}{4x} = \frac{\partial F}{\partial y} . \frac{1}{10y}$

So
$$\frac{\partial F}{\partial x} . \frac{1}{4x} = \frac{\partial F}{\partial y} . \frac{1}{10y}$$
$$\frac{\partial F}{\partial x} . 10y=\frac{\partial F}{\partial y} . 4x$$
$$10y . \frac{\partial F}{\partial x}-4x . \frac{\partial F}{\partial y}=0$$

Then
$$\frac{g(x)}{h(y)}=-\frac{4x}{10y}$$

But the answer is $\frac{4x}{10y}$. Where is my mistake?

Thanks

 

[anuttarasammyak]

I agree with your minus result.

 

[Office_Shredder]

I agree with the minus sign. I disagree with your approach. You demonstrated that $h(y)=10y$ and $g(x)=4x$ happen to satisfy the equation, and then computed the ratio of them. But that doesn't really prove that ratio works for any choice of $g$ and$h$ that satisfy the equation (of which there are many)

I think what they wanted you to do is simply algebra

$h(y)F_x+g(x)F_y=0$
$g(x)F_y=-h(y)F_x$
$g(x)/h(y)=-F_x/F_y$

 

[songoku]

I agree with the minus sign. I disagree with your approach. You demonstrated that $h(y)=10y$ and $g(x)=4x$ happen to satisfy the equation, and then computed the ratio of them. But that doesn't really prove that ratio works for any choice of $g$ and$h$ that satisfy the equation (of which there are many)

I think what they wanted you to do is simply algebra

$h(y)F_x+g(x)F_y=0$
$g(x)F_y=-h(y)F_x$
$g(x)/h(y)=-F_x/F_y$

Ah I see it is possible to start from there. I thought I had to start from partial derivative and tried to rearrange the equation to fit the question.

Thank you very much anuttarasammyak and Office_Shredder