- #1

songoku

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- 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

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

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

##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