# Homework Help: Please check my work (exact equation, then i solved)

1. Apr 15, 2010

### darryw

1. The problem statement, all variables and given/known data
(y-x^3) + (x+y^3)y' = 0

equation is convenienty already in the required form, that is M_(x,y) + N_x(x,y)dy/dx = 0

so..

M_y = 1 = N_x therefore equation is exact. Therefore I now solve...

I am solving for a function, f(x,y) whose partial derivative with respect to y = M
and whose partial with respect to x = N.
Is this correct?

assuming correct so far, my next step is to integrate the M term, with respect to x

integ M_x (y-x^3) = -(1/4)x^4 + h(y)

if y's are treated as constants when integrating, wouldn't the y in this equation become yx?

so that integ_x (y-x^3) = yx -(1/4)x^4 + h(y)
??

assuming yx is not in the integrated result, my next step is to take derivative of the just integrated M term, now with respect to y:

derivative of just-integ M_y [-(1/4)x^4 + h(y) ] = h'(y)

so i know that h'(y) = N term which is:

h'(y) = (x+y^3)

But i think im going to stop here because it looks sort of weird and i want to make sure its right so far.
thanks for any help

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 15, 2010

### Char. Limit

First, yes, the y in your equation becomes yx. This is true.

Second, now that you know this, keep going! You almost have it, I can tell.

3. Apr 15, 2010

### darryw

ok so slight rework to fix the integration error...

integ M_x (y-x^3) = yx -(1/4)x^4 + h(y)

derivative of just-integ M_y [yx -(1/4)x^4 + h(y) ] = x + h'(y)

x + h'(y) = (x+y^3) (the N term)

h'(y) = y^3

h(y) = (1/4)y^4

correct so far? thank you

4. Apr 15, 2010

### darryw

so, final answer: f(x,y) = x + (1/4)y^4

??

5. Apr 15, 2010

### Char. Limit

Well, I've cleaned up your answers for latex format, and it looks like your final answer is missing a term. Just look at the first two of the quotes by you and the answer should be right there.

6. Apr 15, 2010

### darryw

for some reason im confused as to where final answer comes from:
Is it simply N term + whatever I worked out h(y) to be?

If so, answer should be f(x,y) = x + y^3 + (1/4)y^4.

Is this correct and also is answer: N-term + h(y)?
thanks

7. Apr 16, 2010

### Staff: Mentor

The answer should be f(x, y) = C, where f(x, y) is the function you found by integrating M and N. I haven't checked this work, but the idea is exactly the same as in the other thread you posted.

8. Apr 16, 2010

### Char. Limit

Not quite.

The answer is $\int \psi_x dx = \psi + h(y)$

where psi_x is the partial derivative of psi, which means f(x,y).

You found psi, and you found h(y). Those were the first two items I quoted in my last post.

9. Apr 16, 2010

### Staff: Mentor

Since psi means f(x,y), then why not use f(x,y) instead of introducing another symbol that muddies the waters and cannot be typed on standard keyboards?

10. Apr 16, 2010

### Char. Limit

Because I saw psi used as the symbol for f(x,y), and assumed it was common.

Also, how do you write the partial derivative of f(x,y) with respect to x?

Is it ${f(x,y)}_x$?

11. Apr 16, 2010

### Staff: Mentor

The simplest ways use subscripts - fx or fx(x, y).

Or you can use this notation
$$\frac{\partial f}{\partial x}$$

12. Apr 16, 2010

### Char. Limit

Ah.

Just to clear things up, let me resay what I said before, but in more standard notation:

$\int f_x dx = f(x,y) + h(y)$

Thanks for clearing that up, Mark. I'll try to be less confusing from now on.

13. Apr 16, 2010

### Staff: Mentor

Not a problem.