# I need help solving this differential equation

1. Feb 18, 2010

### nikk834

Given that,
f(x,y)= x^3 + 3x^2y + 4xy^2 +2y^3

Prove that x df/dx + y df/dy = 3f

I need help at working this out. I have been trying to figure it out all day.

2. Feb 18, 2010

### pbandjay

First, are you able to calculate ∂f/∂x, ∂f/∂y and 3f?

3. Feb 18, 2010

### nikk834

Hi,
I am not too sure . I have tried to differentiate it but got mixed up variables.

4. Feb 18, 2010

### nikk834

I differentiated df/dx and got 3x^2 + 6xy +4y^2 and for
df/dy i got 3x^2 + 8xy + 6y^2.
Is this right?

5. Feb 18, 2010

### pbandjay

Since f is a function of two variables, x and y, the derivatives of f with respect to x or y are called "partial derivatives", and denoted ∂f/∂x rather than df/dx, for example.

To calculate a partial derivative, then all you have to do is differentiate with respect to the desired variable, viewing all others as constant.

For example: ∂f/∂x = 3x2 + 6xy + 4y2. The "y" is treated as constant in the partial derivative of f with respect to x.

Now, all there is to do is find the partial of f with respect to y and plug them into the equation and see if the equality holds.

6. Feb 18, 2010

### pbandjay

Looks like we're posting at the same time. Yes, those are what I got as well.

7. Feb 18, 2010

### nikk834

Ok.
I found the partial of f with respect to y. so now i got 2 equations, one with respect to x and one with respect to y. what do you plug into the equations exactly and what is 3f?

8. Feb 18, 2010

### pbandjay

Your equation is x (∂f/∂x) + y (∂f/∂y) = 3f. You want to plug your derivatives in for the respective functions in the equation (and distribute the x and y), which will give you the left side of the equation. One the right, plug in the x^3 + 3x^2y + 4xy^2 +2y^3 for f and distribute the 3 across it. Both sides should be equal.

9. Feb 18, 2010

### nikk834

Can you show me what you mean in numbers instead of words because i am having trouble understanding

10. Feb 18, 2010

### pbandjay

Sure, you have x ∂f/∂x + y ∂f/∂y = 3f.

Also, ∂f/∂x = 3x2 + 6xy + 4y2 and ∂f/∂y = 3x2 + 8xy + 6y2.

On the left side of the equation:
x(∂f/∂x) + y(∂f/∂y) = x(3x2 + 6xy + 4y2) + y(3x2 + 8xy + 6y2) = 3x3 + 9x2y + 12xy2 + 6y3

On the right side:
3f = 3(x3 + 3x2y + 4xy2 +2y3) = 3x3 + 9x2y + 12xy2 + 6y3.

Both are equal.

11. Feb 18, 2010

### nikk834

Ok.

12. Feb 19, 2010

### epenguin

Isn't this Euler's theorem? (one of them).

The polynomial is homogeneous. That means all the terms are of the form aixiyn-i.

Not difficult to prove that for any homogeneous polynomial of degree n, f = $$\Sigma$$ aixiyn-i

x (∂f/∂x) + y (∂f/∂y) = nf

13. Feb 20, 2010

### samreen

hey, this is simple eulers theorem application. that is, for a 2 variable fxn,if f(x,y) is a fxn homogeneous of degree n, xdf/dx + ydf/dy = nf(x.y) (note, its all partial derivatives here. so it should be a del operator notation rather than d).

all u need to do is, substitute x=xt and y=yt in the fxn. u will get l.h.s = t^3f(x,y). the degree of t will give the degree of the eqn, if at all homogeneous. and by euler's theorem, the result follows.